6.5 Surfaces, area and flux

A surface is the image of a rectangle or disc under a map r(u, v); the cross product of its two tangent vectors is a normal whose length is the stretching factor from parameter area to surface area, and that one vector gives the area of the surface, the integral of a scalar over it, and the flux of a vector field through it.

A wire is a curve and has one parameter. A sheet of metal, a soap film or the skin of a balloon is a surface and has two. This chapter answers three questions about a surface. How big is it? What is the total of some quantity spread over it, such as mass? And how much of a flow, such as water or heat, crosses it? One vector, built from two tangent vectors, answers all three, and it is the object that Stokes’ theorem and the divergence theorem in 6.6 are written with.

Before you start

2.7: a $2\times 2$ determinant is $ad - bc$ , and a $3\times 3$ one is expanded along a row. 2.9: two vectors are orthogonal when their dot product is $0$ . 5.2: a partial derivative treats the other variables as constants. 6.1: the line integral, and the idea of an oriented curve. 6.3 and 6.4: double integrals over a region, polar coordinates, and the change of variables.

Quick checkWhat is

\det\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}

Quick checkAre

(1, 2, 2)

and

(2, 1, -2)

orthogonal?

Quick checkFor

r(u, v) = (u\cos v,\ u\sin v,\ u^2)

, what is

\dfrac{\partial r}{\partial v}

The picture: patches, normals and flux

Start with the paraboloid and $F = (x, y, z)$ . Slide include strips and watch the green patches fill in: the flux so far is climbing. Each short orange arrow is the vector $N = r_u \times r_v$ of one patch, drawn with length $1$ . Now tick flip the normal: every sign changes, and so does the total. Change the field to $(0, 0, 1)$ and the cone: the flux comes out as $4\pi$ , the area of the disc under the cone, because a field that points straight up only sees the shadow of the surface. Choose the Moebius strip last and read the sentence under the picture.

The idea, in words

A curve is traced by one number $t$ , and $r'(t)$ is the velocity. A surface is traced by two numbers $(u, v)$ . Fix $v$ and move $u$ : the point slides along a curve on the surface with velocity $r_u$ . Fix $u$ and move $v$ : it slides along another curve with velocity $r_v$ . A tiny square of the parameter plane with sides $du$ and $dv$ is sent to a tiny patch of the surface, and the patch looks like a parallelogram with sides $r_u\,du$ and $r_v\,dv$ . The area of that parallelogram is the cross product’s length $\lVert r_u \times r_v \rVert\,du\,dv$ . Add up the patches and you have the area. Put a number on each patch and add, and you have a scalar surface integral. Put a field on each patch and ask how much of it goes through the parallelogram, and you have the flux.

Why· Why a cross product, and why the same vector three times

Think of a rubber sheet stretched over a frame. The parameter square is the unstretched sheet. The vector $r_u \times r_v$ records two things about how the sheet is stretched at one point: how much (its length) and which way the stretched patch faces (its direction, perpendicular to the patch). Area needs only the first. A scalar quantity such as density needs the first. A flow needs both, because the part of a flow that crosses a patch depends on how the patch is tilted against the flow, and a patch lying along the flow lets nothing through. So one vector, $N = r_u \times r_v$ , serves all three jobs, and the entire chapter is learning to compute it and to use it with the right sign.

Where this came from· Why area of a surface was harder than length of a curve

You can find the length of a curve by laying a polygon with corners on the curve and letting the corners get closer together. In the 1860s a standard French textbook, Serret’s Cours de calcul différentiel et intégral, said the area of a curved surface is likewise the limit of the areas of inscribed polyhedra. Around 1880 Hermann Schwarz found a cylinder for which this fails: stack thin rings of triangles with the rings twisted against each other (the shape is called Schwarz’s lantern) and the area of the polyhedron can approach a number larger than the cylinder’s area, or grow without bound, even though every corner lies on the cylinder. Peano found the same example independently around 1882. The reason is that very thin, tilted triangles can hug the cylinder at their corners while leaning far away from its tangent planes. So a definition of area must use tangent planes, not only points, and that is exactly what the fundamental vector product does. The lab in the area section runs the lantern.

The cross product

Everything below needs one operation on two vectors in space. It is worth building it from what it does, because the formula is easy to misremember.

Definition T

Cross product

The cross product of $a$ and $b$ is the vector

a \times b = (a_2b_3 - a_3b_2,\ \ a_3b_1 - a_1b_3,\ \ a_1b_2 - a_2b_1).

It is remembered as the determinant of a $3\times 3$ table with $e_1 = (1,0,0)$ , $e_2 = (0,1,0)$ , $e_3 = (0,0,1)$ in the first row, $a$ in the second and $b$ in the third, expanded along the first row (2.7):

a \times b = \det\begin{bmatrix} e_1 & e_2 & e_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{bmatrix} = e_1(a_2b_3 - a_3b_2) - e_2(a_1b_3 - a_3b_1) + e_3(a_1b_2 - a_2b_1).

The middle term carries a minus sign from the cofactor expansion, which is why the middle component of the formula reads $a_3b_1 - a_1b_3$ and not the other way round. Unlike the dot product, it needs three dimensions, and the result is a vector, not a number.

In one lineFor $a = (a_1, a_2, a_3)$ and $b = (b_1, b_2, b_3)$ in $\mathbb R^3$ , $a \times b = (a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1)$ : a vector perpendicular to both, whose length is the area of the parallelogram with sides $a$ and $b$ .

Why it existsArea and direction of a tilted patch in one vector. Both are needed to measure a surface and to say which way it faces.

Why· Where the formula comes from

The lecture does not start from the formula. It starts from what the cross product must do: it is linear in each slot, $a\times b = -\,b\times a$ (so $a\times a = 0$ ), and $e_1\times e_2 = e_3$ , $e_2\times e_3 = e_1$ , $e_3\times e_1 = e_2$ . Write $a = \sum a_ie_i$ and $b = \sum b_je_j$ and multiply out using linearity: $a\times b = \sum_{i,j}a_ib_j\,e_i\times e_j$ . The three terms with $i = j$ are $0$ . Pair the term $(i, j)$ with $(j, i)$ , which has the opposite sign, to get $\sum_{i<j}(a_ib_j - a_jb_i)\,e_i\times e_j$ . The pair $(1, 2)$ gives $(a_1b_2 - a_2b_1)\,e_3$ , the pair $(2, 3)$ gives $(a_2b_3 - a_3b_2)\,e_1$ , and the pair $(1, 3)$ gives $(a_1b_3 - a_3b_1)\,e_1\times e_3 = (a_3b_1 - a_1b_3)\,e_2$ , because $e_1\times e_3 = -e_3\times e_1 = -e_2$ . That last minus sign is the flipped middle component.

Theorem T MATH 376 lectures of 23 and 25 February

Length and direction of the cross product

Let $\theta \in [0, \pi]$ be the angle between $a$ and $b$ . Then the facts in the statement hold. The sign rule $e_1\times e_2 = e_3$ , $e_2\times e_3 = e_1$ , $e_3\times e_1 = e_2$ is the right-hand rule: curl the fingers of your right hand from $a$ towards $b$ and the thumb points along $a\times b$ .

In one line $a \times b$ is orthogonal to $a$ and to $b$ ; $\lVert a \times b\rVert^2 = \lVert a\rVert^2\lVert b\rVert^2 - (a\cdot b)^2$ ; so $\lVert a\times b\rVert = \lVert a\rVert\lVert b\rVert\sin\theta$ is the area of the parallelogram with sides $a$ and $b$ . Also $b \times a = -a \times b$ and $e_1\times e_2 = e_3$ .

Why it is trueIt says what the cross product is for: the direction is perpendicular to the plane of $a$ and $b$ and the length is the area, so the formula is the right vector to measure and to orient a patch.

ProofMATH 376 lecture of 25 February

The idea: Perpendicularity is a dot product that cancels term by term. The length comes from an identity that both sides expand to the same six products; then the formula for the dot product in terms of the angle turns it into the area of a parallelogram: base times height.

a\cdot(a\times b) = a_1(a_2b_3 - a_3b_2) + a_2(a_3b_1 - a_1b_3) + a_3(a_1b_2 - a_2b_1)

Dot the formula with

a

and multiply out.

= a_1a_2b_3 - a_1a_3b_2 + a_2a_3b_1 - a_1a_2b_3 + a_1a_3b_2 - a_2a_3b_1 = 0

Collect: each of the six products appears once with a plus and once with a minus.Doing the same with

b

shows it is perpendicular to

b

. A cross product that fails these two dot products has a sign error in it.

\lVert a\times b\rVert^2 = \sum_{i\ne j} a_i^2b_j^2 - 2\,(a_1a_2b_1b_2 + a_1a_3b_1b_3 + a_2a_3b_2b_3)

, where the sum has the six terms with

i \ne j

Square the three components. Each square gives two squared terms and one cross term.

\lVert a\rVert^2\lVert b\rVert^2 = \sum_i a_i^2b_i^2 + \sum_{i\ne j}a_i^2b_j^2

and

(a\cdot b)^2 = \sum_i a_i^2b_i^2 + 2\,(a_1a_2b_1b_2 + a_1a_3b_1b_3 + a_2a_3b_2b_3)

Multiply out the product of the two squared lengths, and the square of the dot product.

\lVert a\rVert^2\lVert b\rVert^2 - (a\cdot b)^2 = \lVert a\times b\rVert^2

Subtract. The terms

\sum_i a_i^2b_i^2

cancel and what is left is the line above.

\lVert a\times b\rVert^2 = \lVert a\rVert^2\lVert b\rVert^2\sin^2\theta

The dot product is

a\cdot b = \lVert a\rVert\lVert b\rVert\cos\theta

(2.9), and

1 - \cos^2\theta = \sin^2\theta

\lVert a\times b\rVert = \lVert a\rVert\lVert b\rVert\sin\theta

Take square roots.

\sin\theta \ge 0

[0, \pi]

, so no sign is needed.

\lVert a\times b\rVert

is the area of the parallelogram.

\blacksquare

A parallelogram has area base times height: the base is

\lVert a\rVert

and the height is

\lVert b\rVert\sin\theta

. Swapping

a

and

b

flips the sign of each component in the formula.The sign claim

b \times a = -a\times b

is the same swap:

b_2a_3 - b_3a_2 = -(a_2b_3 - a_3b_2)

, and likewise for the other two components.

Proof workshopRebuild the proof that the length of a x b is the parallelogram areaMATH 376 lecture of 25 February

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Worked example MATH 376 worksheet 12 problems 1 and 2

Compute

(2, -1, 3) \times (1, 4, -2)

, check that it is orthogonal to both vectors, and find the area of the parallelogram and of the triangle they span.

Solution.

First component. $a_2b_3 - a_3b_2 = (-1)(-2) - (3)(4) = 2 - 12 = -10$ . Rows $2$ and $3$ of the pattern: ignore $a_1$ and $b_1$ .
Second component. $a_3b_1 - a_1b_3 = (3)(1) - (2)(-2) = 3 + 4 = 7$ . Remember the flipped order: third times first, minus first times third.
Third component. $a_1b_2 - a_2b_1 = (2)(4) - (-1)(1) = 8 + 1 = 9$ . Ignore $a_3$ and $b_3$ .
Check with two dot products. $(-10, 7, 9)\cdot(2, -1, 3) = -20 - 7 + 27 = 0$ and $(-10, 7, 9)\cdot(1, 4, -2) = -10 + 28 - 18 = 0$ . A cross product must be orthogonal to both factors. This catches the usual sign slip in the middle component.
Length. $\lVert(-10, 7, 9)\rVert = \sqrt{100 + 49 + 81} = \sqrt{230}$ . The length of $a\times b$ is the parallelogram area.
Triangle. The triangle with sides $a$ and $b$ is half the parallelogram, so its area is $\frac12\sqrt{230}$ .

Answer. $a\times b = (-10, 7, 9)$ ; parallelogram area $\sqrt{230}\approx 15.17$ ; triangle area $\frac12\sqrt{230}\approx 7.58$ . Worksheet 12 problem 2 is $(1, 2, 2)\times(2, 0, 1) = (2, 3, -4)$ , and the same two dot products give $2 + 6 - 8 = 0$ and $4 + 0 - 4 = 0$ .

Worked example MATH 376 worksheet 12 problem 3

Find a normal vector to the plane through

A = (1, 0, 2)

B = (3, -1, 1)

C = (2, 2, 0)

, and the area of the triangle

ABC

Solution.

Two edges from one corner. $\overrightarrow{AB} = B - A = (2, -1, -1)$ and $\overrightarrow{AC} = C - A = (1, 2, -2)$ . Both lie in the plane, and they are not parallel.
Cross them. $\overrightarrow{AB}\times\overrightarrow{AC} = ((-1)(-2) - (-1)(2),\ (-1)(1) - (2)(-2),\ (2)(2) - (-1)(1)) = (4, 3, 5)$ . The cross product is orthogonal to two vectors in the plane, so it is a normal to the plane.
Area. $\frac12\lVert(4, 3, 5)\rVert = \frac12\sqrt{16 + 9 + 25} = \frac12\sqrt{50} = \frac{5\sqrt2}{2}$ . Half the parallelogram.

Answer. A normal is $(4, 3, 5)$ and the area is $\frac{5\sqrt2}{2}\approx 3.54$ . The plane is $4(x - 1) + 3y + 5(z - 2) = 0$ .

Your answerFind the area of the triangle with vertices

A = (1, 1, 0)

B = (3, 1, 2)

and

C = (1, 4, 2)

Quick checkWhat is

(1, 2, 3)\times(2, 4, 6)

Quick checkYou compute

a\times b

and get a vector whose dot product with

a

is not

0

. What happened?

Parametric surfaces

Definition T

Parametric surface

A curve needs one parameter, running over an interval. A surface needs two, running over a region $T$ of the plane, such as a rectangle or a disc. The map $r: T \to \mathbb R^3$ is a parametrization and $S = r(T)$ is a parametric surface. When $r(x, y) = (x, y, g(x, y))$ the parametrization is explicit: $S$ is the graph $z = g(x, y)$ . The number of parameters is the dimension: a circle and a helix have one, a disc (even though it is flat), a cylinder and a sphere have two.

In one lineA surface $S$ given by a map $r(u, v) = (x(u, v),\ y(u, v),\ z(u, v))$ from a parameter region $T$ in the $uv$ -plane; $S$ is the set of points $r(u, v)$ with $(u, v)$ in $T$ .

Why it existsIt turns a two-dimensional object in space into two ordinary variables, so every question about the surface becomes a double integral over a flat region.

Memorize this table. Each row is used all through the chapter and the next.

Surface	Parametrization	Parameter region
graph $z = g(x, y)$	$r(x, y) = (x,\ y,\ g(x, y))$	the region $D$ of the $xy$ -plane under it
cylinder $x^2 + y^2 = R^2$	$r(\theta, z) = (R\cos\theta,\ R\sin\theta,\ z)$	$0 \le \theta \le 2\pi$ , $z$ in an interval
cone $z = \sqrt{x^2 + y^2}$	$r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ r)$	$r \ge 0$ , $0 \le \theta \le 2\pi$
paraboloid $z = 4 - x^2 - y^2$	$r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ 4 - r^2)$	$0 \le r \le 2$ , $0 \le \theta \le 2\pi$
sphere of radius $R$	$r(\varphi, \theta) = R(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$	$0 \le \varphi \le \pi$ , $0 \le \theta \le 2\pi$
ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$	$r(u, v) = (a\sin v\cos u,\ b\sin v\sin u,\ c\cos v)$	$0 \le u \le 2\pi$ , $0 \le v \le \pi$
torus: circle of radius $a$ swept round a circle of radius $R > a$	$r(u, v) = ((R + a\cos v)\cos u,\ (R + a\cos v)\sin u,\ a\sin v)$	$0 \le u, v \le 2\pi$
plane through $P$ with sides $a, b$	$r(u, v) = P + ua + vb$	$0 \le u \le 1$ , $0 \le v \le 1$ for the parallelogram

On the sphere, $\varphi$ is the angle measured down from the north pole (the positive $z$ -axis) and $\theta$ is the angle around the $z$ -axis. In the cone and paraboloid rows the letter $r$ is both the radius coordinate and the name of the map; the name of the map is $r(\cdot,\cdot)$ with two arguments.

Watch out· The triangle is not the parallelogram

The set $\{ua + vb : 0\le u, v\le 1\}$ is the parallelogram. The set with the extra condition $u + v \le 1$ is only the triangle with corners $0$ , $a$ , $b$ . A wrong parameter region gives a different set, and a different answer.

Worked example MATH 376 worksheet 11 problems 1 to 4

Parametrize the unit sphere, the cylinder

x^2 + y^2 = 9

with

0 < z < 5

, the slanted triangle

x + 2y + z = 2

in the first octant, and the punctured disc

z = x

with

0 < x^2 + y^2 < 1

. In each case find the infimum and supremum of the coordinate

x

and say whether they are attained.

The infimum of the values of $x$ is the largest number that is at most every value, the supremum is the smallest number that is at least every value, and a bound is attained if some point of the surface has exactly that value of $x$ .

Solution.

Sphere. $r(\varphi, \theta) = (\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$ , $0\le\varphi\le\pi$ , $0\le\theta\le2\pi$ . Then $x = \sin\varphi\cos\theta$ has $\lvert x\rvert \le 1$ , and $\varphi = \pi/2$ , $\theta = 0$ gives $x = 1$ , while $\theta = \pi$ gives $x = -1$ . Both bounds are reached, so the supremum $1$ and infimum $-1$ are attained.
Cylinder. $r(\theta, z) = (3\cos\theta,\ 3\sin\theta,\ z)$ , $\theta\in[0, 2\pi)$ , $z\in(0, 5)$ . Then $x = 3\cos\theta\in[-3, 3]$ , and $\theta = 0$ and $\theta = \pi$ reach both ends for any $z$ . Attained. The missing end circles $z = 0$ and $z = 5$ do not matter for $x$ . For the coordinate $z$ instead, the infimum $0$ and the supremum $5$ would not be attained.
Triangle. The plane meets the axes at $(2, 0, 0)$ , $(0, 1, 0)$ , $(0, 0, 2)$ . Solve for $z$ : $r(u, v) = (u,\ v,\ 2 - u - 2v)$ with $0\le u\le2$ , $0\le v\le 1 - u/2$ (this is $z \ge 0$ ). Then $x = u\in[0, 2]$ , with $u = 2$ at the vertex $(2, 0, 0)$ and $u = 0$ along the edge $x = 0$ . Both are attained: the triangle contains its edges.
Punctured disc. $r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ r\cos\theta)$ with $0 < r < 1$ , $0\le\theta\le2\pi$ . Then $\lvert x\rvert\le r < 1$ with values arbitrarily close to $1$ and $-1$ (take $\theta = 0$ or $\pi$ and $r\to1$ ). The bound $1$ would need $r = 1$ , which was removed, so the supremum $1$ and the infimum $-1$ are not attained.

Answer. Sphere: $\pm1$ , attained. Cylinder: $\pm3$ , attained. Triangle: $0$ and $2$ , attained. Punctured disc: $\pm1$ , not attained. (The worksheet calls a surface ‘closed’ when it has no boundary curve, like the sphere, and ‘open’ when it has one. That is a different use of the words from the point-set words in 4.1.)

Worked example MATH 376 worksheet 15 problem 1

Parametrize each of: a circle of radius

R

, a disc of radius

R

, a helix of radius

R

and one turn of height

H

, and a cylinder of radius

R

and height

h

. How many parameters does each need?

Solution.

Circle. $(R\cos t,\ R\sin t)$ , $0\le t\le2\pi$ : one parameter, an interval.
Disc. $(r\cos\theta,\ r\sin\theta)$ , $0\le r\le R$ , $0\le\theta\le2\pi$ : two parameters. The disc is flat and lives in the plane, but it is a surface: it takes two numbers to name a point.
Helix. $(R\cos t,\ R\sin t,\ Ht/(2\pi))$ , $0\le t\le2\pi$ : one parameter. It is a curve in space.
Cylinder. $(R\cos\theta,\ R\sin\theta,\ z)$ , $0\le\theta\le2\pi$ , $0\le z\le h$ : two parameters.

Answer. One, two, one, two. A curve has a one-dimensional parameter set and a surface a two-dimensional one, whatever space they sit in.

The fundamental vector product

Fix a point $(u_0, v_0)$ of the parameter region. The partial derivatives $r_u = \partial r/\partial u$ and $r_v = \partial r/\partial v$ are two vectors at the point $r(u_0, v_0)$ . Moving $u$ by a little bit $\delta$ moves the point by about $\delta\,r_u$ (the linear approximation of 5.3), and moving $v$ by $\delta$ moves it by about $\delta\,r_v$ . So the small square $[u_0, u_0 + \delta]\times[v_0, v_0+\delta]$ is sent to a patch that is, to first order, the parallelogram with sides $\delta\,r_u$ and $\delta\,r_v$ . By the theorem above, its area is $\lVert r_u\times r_v\rVert\,\delta^2$ .

Definition T

Fundamental vector product

The fundamental vector product of $r$ at $(u, v)$ is

N(u, v) = r_u \times r_v.

The parametrization is regular at $(u, v)$ if $N(u, v) \ne 0$ , and smooth if it is regular at every point (and $r$ is continuously differentiable). Where $N = 0$ the two tangent vectors are parallel or one of them is zero, the patch collapses to a segment or a point, there is no tangent plane and no unit normal.

In one lineFor a parametrization $r(u, v)$ the vector $N(u, v) = r_u \times r_v$ . It is normal to the surface and $\lVert N\rVert$ is the area of the image of a unit parameter square. The parametrization is regular where $N \ne 0$ and smooth if it is regular everywhere.

Why it existsIt is the one vector that carries the facing direction and the stretching factor of a surface patch, so every surface integral is written with it.

Theorem T MATH 376 lectures of 23 to 27 February; Apostol 12.1 to 12.5

Normal and area element of a parametrized surface

At a regular point, $N$ is orthogonal to $r_u$ and $r_v$ , which span the tangent plane, so it is a normal vector to the surface. The unit normal is $n = N/\lVert N\rVert$ . A small parameter rectangle of area $du\,dv$ is sent to a patch of area $\lVert N\rVert\,du\,dv$ to first order, so we write $dS = \lVert N\rVert\,du\,dv$ .

In one line $N = r_u \times r_v$ is perpendicular to the tangent plane at each regular point, and the area element is $dS = \lVert r_u\times r_v\rVert\,du\,dv$ .

Why it is trueThis is the whole reason the cross product enters surface integrals: its direction is the normal and its length is the area stretching factor.

The theorem is the computation above: the first sentence is the perpendicularity part of the cross product theorem, and the second is its length part applied to the patch with sides $r_u\,du$ and $r_v\,dv$ . The ideas of linear approximation in 5.3 are what turn “to first order” into a limit, and it is the same limit that defines the surface area in the next section.

Proposition T MATH 376 lecture of 2 March and worksheet 12; Apostol 12.5

Normal of a graph

For a graph of a continuously differentiable function $g$ , the parametrization $r(x, y) = (x, y, g(x, y))$ has the stated fundamental vector product, and it is regular everywhere.

In one lineFor the graph $z = g(x, y)$ with $r(x, y) = (x, y, g(x, y))$ : $r_x \times r_y = (-g_x, -g_y, 1)$ , which points upward, and $\lVert r_x\times r_y\rVert = \sqrt{1 + g_x^2 + g_y^2}$ .

Why it is trueGraphs are the most common surfaces, and this formula turns most flux and area problems about them into a one-line computation.

ProofMATH 376 lecture of 2 March

The idea: Differentiate, then cross. The first two components of the cross product each have one zero factor, and the third is a determinant of a matrix with $1, 0, 0, 1$ in it.

r_x = (1, 0, g_x)

and

r_y = (0, 1, g_y)

Differentiate each coordinate of

r(x, y) = (x, y, g(x, y))

with respect to

x

and then

y

r_x\times r_y = (0\cdot g_y - g_x\cdot1,\ \ g_x\cdot0 - 1\cdot g_y,\ \ 1\cdot1 - 0\cdot0)

The cross product formula with

a = r_x

and

b = r_y

= (-g_x,\ -g_y,\ 1)

Evaluate each component.The third component is

+1

for every graph: that is why

r_x\times r_y

always points up, and why we say 'upward' for a graph without checking anything else.

\lVert r_x\times r_y\rVert = \sqrt{g_x^2 + g_y^2 + 1}

\blacksquare

Length of a vector: square root of the sum of squares. It is at least

1

, so never zero.

Why· The same fact seen from the side

The factor $\sqrt{1 + g_x^2 + g_y^2}$ equals $1/\cos\gamma$ , where $\gamma$ is the angle between the upward normal and the vertical. A steep patch has a large $\gamma$ and its true area is its shadow on the $xy$ -plane divided by $\cos\gamma$ : a roof of slope $1$ ( $g_x = 1$ , $g_y = 0$ ) has $\sqrt2$ times the area of the floor under it. This is also why a flat piece of paper held at an angle shades less of the table than it covers.

Worked example MATH 376 worksheet 11 problem 5

Find the tangent vectors and the fundamental vector product of

r(u, v) = (u^2,\ \sin(uv),\ e^v)

(u, v) = (1, 0)

Solution.

Differentiate in $u$ . $r_u = (2u,\ v\cos(uv),\ 0)$ . Chain rule on $\sin(uv)$ : the derivative of $uv$ in $u$ is $v$ . The last coordinate has no $u$ .
Differentiate in $v$ . $r_v = (0,\ u\cos(uv),\ e^v)$ .
Evaluate at $(1, 0)$ . $r_u = (2, 0, 0)$ and $r_v = (0, 1, 1)$ . $v\cos(uv) = 0$ and $u\cos(uv) = \cos 0 = 1$ , $e^0 = 1$ .
Cross. $N = (0\cdot1 - 0\cdot1,\ 0\cdot0 - 2\cdot1,\ 2\cdot1 - 0\cdot0) = (0, -2, 2)$ .
Check. $N\cdot r_u = 0$ and $N\cdot r_v = -2 + 2 = 0$ . The free check from the cross product theorem.

Answer. $r_u = (2, 0, 0)$ , $r_v = (0, 1, 1)$ , $N = (0, -2, 2)$ .

Worked example new (compare 5.3)

Find the equation of the tangent plane to the paraboloid

z = x^2 + y^2

(1, 1, 2)

from the fundamental vector product.

Solution.

Normal of a graph. $g = x^2 + y^2$ has $g_x = 2x = 2$ and $g_y = 2y = 2$ at the point, so $N = (-2, -2, 1)$ . The graph formula.
Plane through the point with this normal. $-2(x - 1) - 2(y - 1) + (z - 2) = 0$ , that is $z = 2x + 2y - 2$ . The tangent plane is the set of points $X$ with $N\cdot(X - p) = 0$ .
Check against 5.3. The level-surface gradient of $f = x^2 + y^2 - z$ is $(2x, 2y, -1) = (2, 2, -1)$ , which is $-N$ . Same line, opposite direction: both are normals, so both describe the same plane.

Answer. $z = 2x + 2y - 2$ .

Worked example MATH 376 Midterm 2 problem 1(a); see [A.3](/vol2/a3-midterm-2/)

Find the normal field of the surface

S = \{(u, v, e^u) : 0 \le u, v \le 1\}

pointing in the positive

z

direction.

Solution.

Parametrize. $r(u, v) = (u, v, e^u)$ . It is the graph of $z = e^x$ over the unit square, with $u = x$ and $v = y$ .
Tangent vectors. $r_u = (1, 0, e^u)$ and $r_v = (0, 1, 0)$ .
Cross. $N = r_u\times r_v = (0\cdot0 - e^u\cdot1,\ e^u\cdot0 - 1\cdot0,\ 1\cdot1 - 0\cdot0) = (-e^u, 0, 1)$ . The third component is $+1 > 0$ , so it points in the positive $z$ direction as the problem asks. No sign flip.
Length. $\lVert N\rVert = \sqrt{e^{2u} + 1}$ .
Normalize. $n = \dfrac{(-e^u,\ 0,\ 1)}{\sqrt{e^{2u} + 1}}$ . The unit normal is $N$ divided by its length.

Answer. $n = (-e^u, 0, 1)/\sqrt{e^{2u} + 1}$ . The practice exam asks the same for $(u, v, v - u^3)$ (A.5), where $r_u = (1, 0, -3u^2)$ , $r_v = (0, 1, 1)$ and $N = (3u^2, -1, 1)$ come out the same way.

Quick checkFor the surface

r(u, v) = (u, v, u^2 + v^2)

, which is the fundamental vector product?

Surface area

Definition T

Surface area

For a smooth parametrization $r: T\to S$ of a surface $S$ , the surface area of $S$ is

\operatorname{area}(S) = \iint_T \lVert r_u\times r_v\rVert\,du\,dv.

For a graph $z = g(x, y)$ over a region $D$ this becomes $\iint_D \sqrt{1 + g_x^2 + g_y^2}\,dx\,dy$ .

In one lineThe area of $S = r(T)$ for a smooth parametrization is $\iint_T \lVert r_u \times r_v\rVert\,du\,dv$ .

Why it existsIt adds up the areas $\lVert N\rVert\,du\,dv$ of the parallelogram patches, and is the only definition that survives the Schwarz lantern.

Method: (1) parametrize, using the constraints of the problem to set the parameter ranges; (2) compute $r_u$ , $r_v$ and $N = r_u\times r_v$ ; (3) take the length; (4) integrate over the parameter region.

Worked example MATH 376 Midterm 2 problem 2, 6 points; see [A.3](/vol2/a3-midterm-2/)

Find the area of the part of the paraboloid

z = \frac34 - x^2 - y^2

above the plane

z = 0

, using cylindrical coordinates. (Hint:

\frac{d}{dr}(1 + 4r^2)^{3/2} = 12\,r\sqrt{1 + 4r^2}

Solution.

Parameter ranges. $z\ge0$ means $\frac34 - r^2\ge0$ , so $0\le r\le\frac{\sqrt3}2$ and $0\le\theta\le2\pi$ . The constraint of the problem is what sets the limits.
Parametrize. $r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ \frac34 - r^2)$ .
Tangent vectors. $r_r = (\cos\theta,\ \sin\theta,\ -2r)$ and $r_\theta = (-r\sin\theta,\ r\cos\theta,\ 0)$ .
Cross. $r_r\times r_\theta = (\sin\theta\cdot0 - (-2r)(r\cos\theta),\ (-2r)(-r\sin\theta) - \cos\theta\cdot0,\ \cos\theta\cdot r\cos\theta + \sin\theta\cdot r\sin\theta) = (2r^2\cos\theta,\ 2r^2\sin\theta,\ r)$ .
Length. $\sqrt{4r^4\cos^2\theta + 4r^4\sin^2\theta + r^2} = \sqrt{4r^4 + r^2} = r\sqrt{1 + 4r^2}$ . $\cos^2 + \sin^2 = 1$ , then pull $r^2$ out of the root; this is allowed because $r\ge0$ .
Integrate in $r$ . By the hint, $\int_0^{\sqrt3/2}r\sqrt{1 + 4r^2}\,dr = \frac1{12}\big[(1 + 4r^2)^{3/2}\big]_0^{\sqrt3/2} = \frac1{12}(4^{3/2} - 1) = \frac7{12}$ . At $r = \sqrt3/2$ : $1 + 4\cdot\frac34 = 4$ and $4^{3/2} = 8$ . At $r = 0$ the value is $1$ . If there were no hint, you would guess a multiple of $(1+4r^2)^{3/2}$ and differentiate it to find the multiple.
Integrate in $\theta$ . The inner integral does not depend on $\theta$ , so multiply by $2\pi$ : $2\pi\cdot\frac7{12} = \frac{7\pi}6$ .

Answer. $\dfrac{7\pi}{6}\approx3.665$ . The flat disc under it has area $\pi\cdot\frac34 = 2.356$ , and the cap is bigger because it is tilted.

Worked example MATH 376 practice Midterm 2 problem 2; see [A.5](/vol2/a5-practice-exams/)

Find the area of the cone

z = \sqrt{x^2 + y^2}

between the planes

z = 0

and

z = 2^{1/4}

, using cylindrical coordinates.

Solution.

Parametrize. $r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ r)$ with $0\le r\le2^{1/4}$ , $0\le\theta\le2\pi$ . On the cone $z = r$ , so $0\le z\le2^{1/4}$ is $0\le r\le2^{1/4}$ .
Cross. $r_r = (\cos\theta, \sin\theta, 1)$ and $r_\theta = (-r\sin\theta, r\cos\theta, 0)$ , so $r_r\times r_\theta = (-r\cos\theta,\ -r\sin\theta,\ r)$ .
Length. $\lVert\cdot\rVert = \sqrt{r^2\cos^2\theta + r^2\sin^2\theta + r^2} = r\sqrt2$ .
Integrate. $\displaystyle\int_0^{2\pi}\!\!\int_0^{2^{1/4}}\sqrt2\,r\,dr\,d\theta = 2\pi\sqrt2\cdot\frac{(2^{1/4})^2}{2} = 2\pi\sqrt2\cdot\frac{\sqrt2}2 = 2\pi$ .
Check with geometry. The lateral area of a cone is $\pi\cdot\text{radius}\cdot\text{slant height} = \pi\cdot2^{1/4}\cdot\sqrt2\cdot2^{1/4} = \pi\sqrt2\cdot\sqrt2 = 2\pi$ . A second method, from high-school geometry, agrees.

Answer. $2\pi$ . At the apex $r = 0$ the vector $N$ is $0$ , so the parametrization is not regular there. The apex is a single point, so it changes no integral.

Worked example Apostol 12.5 (sphere); new (torus)

Find the area of a sphere of radius

R

, and of the torus with radii

R > a

Solution.

Sphere, tangent vectors. $r(\varphi,\theta) = R(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$ has $r_\varphi = R(\cos\varphi\cos\theta,\ \cos\varphi\sin\theta,\ -\sin\varphi)$ and $r_\theta = R(-\sin\varphi\sin\theta,\ \sin\varphi\cos\theta,\ 0)$ .
Sphere, cross. $r_\varphi\times r_\theta = R^2\sin\varphi\,(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$ . The first component is $0 - (-\sin\varphi)(\sin\varphi\cos\theta)R^2 = R^2\sin^2\varphi\cos\theta$ , and similarly for the others. The third is $R^2\sin\varphi\cos\varphi(\cos^2\theta + \sin^2\theta)$ .
Sphere, length. The vector in brackets has length $1$ , and $\sin\varphi\ge0$ on $[0, \pi]$ , so $\lVert N\rVert = R^2\sin\varphi$ .
Sphere, integrate. $\int_0^{2\pi}\!\int_0^\pi R^2\sin\varphi\,d\varphi\,d\theta = 2\pi R^2\cdot2 = 4\pi R^2$ .
Torus. With $r(u, v) = ((R + a\cos v)\cos u,\ (R + a\cos v)\sin u,\ a\sin v)$ the cross product $r_u\times r_v = a(R + a\cos v)\,(\cos u\cos v,\ \sin u\cos v,\ \sin v)$ , which has length $a(R + a\cos v)$ since $R > a$ . Same pattern as the sphere: a scalar times an outward unit vector.
Torus, integrate. $\int_0^{2\pi}\!\int_0^{2\pi}a(R + a\cos v)\,du\,dv = 2\pi\cdot a\cdot(2\pi R + 0) = 4\pi^2Ra$ . $\int_0^{2\pi}\cos v\,dv = 0$ .

Answer. Sphere $4\pi R^2$ . Torus $4\pi^2Ra$ : the length of the core circle, $2\pi R$ , times the circumference of the tube, $2\pi a$ .

Worked example Apostol 12.5 (surfaces of revolution)

Derive the high-school formula

2\pi\int y\sqrt{1 + y'^2}\,dx

for the area of a surface of revolution, and use it to show that the part of a sphere of radius

R

between two parallel planes at distance

h

has area

2\pi Rh

Solution.

Parametrize the surface of revolution. Rotate the graph $y = f(x)\ge0$ , $a\le x\le b$ , about the $x$ -axis: $r(x, \theta) = (x,\ f(x)\cos\theta,\ f(x)\sin\theta)$ .
Cross. $r_x = (1,\ f'\cos\theta,\ f'\sin\theta)$ and $r_\theta = (0,\ -f\sin\theta,\ f\cos\theta)$ , so $r_x\times r_\theta = (ff',\ -f\cos\theta,\ -f\sin\theta)$ . First: $f'\cos\theta\cdot f\cos\theta + f'\sin\theta\cdot f\sin\theta = ff'$ . Second: $f'\sin\theta\cdot0 - 1\cdot f\cos\theta$ . Third: $1\cdot(-f\sin\theta) - 0$ .
Length. $\sqrt{f^2f'^2 + f^2} = f\sqrt{1 + f'^2}$ .
Area. $\int_a^b\int_0^{2\pi}f\sqrt{1 + f'^2}\,d\theta\,dx = 2\pi\int_a^bf\sqrt{1 + f'^2}\,dx$ . That is the formula from JEE: the rule you used is this computation with the angle already integrated.
The sphere. For $f = \sqrt{R^2 - x^2}$ , $f' = -x/f$ , so $f\sqrt{1 + f'^2} = \sqrt{f^2 + x^2} = R$ .
The zone. The area between $x = a$ and $x = b$ is $2\pi R\,(b - a) = 2\pi Rh$ . It does not depend on where the planes cut: slices of equal thickness have equal area (Archimedes).

Answer. The zone has area $2\pi Rh$ ; the whole sphere ( $h = 2R$ ) has area $4\pi R^2$ .

Your answerFind the area of the part of the paraboloid

z = x^2 + y^2

with

z \le 1

Your answerFind the area of the torus whose core circle has radius

3

and whose tube has radius

1

Beyond the course: for understanding· Label B: the lantern and the torus

Schwarz’s lantern, the surface of revolution formula and the torus are beyond MATH 376, which set area problems only for graphs, the cone and the paraboloid cap. They are here to show why the definition of area uses tangent planes and to give more surfaces to practise the cross product on.

Python labArea of a surface by a mesh of triangles, and Schwarz's lantern

import numpy as np

def mesh_area(r, a, b, c, d, n):
    u = np.linspace(a, b, n + 1); v = np.linspace(c, d, n + 1)
    U, V = np.meshgrid(u, v, indexing="ij")
    P = np.stack(r(U, V), axis=-1)
    A, B, C, D = P[:-1, :-1], P[1:, :-1], P[1:, 1:], P[:-1, 1:]
    tri = lambda p, q, s: 0.5 * np.linalg.norm(np.cross(q - p, s - p), axis=-1).sum()
    return tri(A, B, C) + tri(A, C, D)

sphere = lambda ph, th: (np.sin(ph) * np.cos(th), np.sin(ph) * np.sin(th), np.cos(ph))
print("n   mesh area of unit sphere   error")
for n in (4, 8, 16, 32, 64):
    A = mesh_area(sphere, 0, np.pi, 0, 2 * np.pi, n)
    print(f"{n:3d}  {A:.6f}  {abs(A - 4 * np.pi):.6f}")

def lantern(m, n):   # m slabs, n points per ring, neighbouring rings twisted by pi/n
    return 2 * n * m * np.sin(np.pi / n) * np.sqrt(1 / m**2 + (1 - np.cos(np.pi / n))**2)

print("true area of the cylinder:", round(2 * np.pi, 4))
for n in (10, 20, 40):
    print(n, " m=n:", round(lantern(n, n), 4), " m=n^2:", round(lantern(n**2, n), 4), " m=n^3:", round(lantern(n**3, n), 1))

Try this. Run it. The first table cuts the sphere into n by n parameter squares, each into two flat triangles, and adds the triangle areas (half the length of a cross product). The error against 4*pi shrinks by a factor of about 4 each time n doubles, so it falls like 1/n^2 and the mesh area converges. The second part is Schwarz's lantern: a cylinder of radius 1 and height 1, true area 2*pi = 6.2832, replaced by a polyhedron with m rings of n points, every corner on the cylinder. With m = n the area goes to 2*pi. With m = n^2 it goes to about 31.6, five times too big. With m = n^3 it grows without bound. All three have more and more triangles and every corner on the cylinder. Change m to 2*n or to n**2.5 and predict before you run.

Quick checkA cylinder of radius

3

is parametrized by

r(\theta, z) = (3\cos\theta, 3\sin\theta, z)

on the rectangle

[0, 2\pi]\times[0, 3]

. Which is the area of the cylinder?

Quick checkThe apex of the cone

r(r, \theta) = (r\cos\theta, r\sin\theta, r)

is the point

r = 0

. What is

r_r\times r_\theta

there?

Integrating a scalar over a surface

Definition T

Surface integral

The surface integral of a continuous scalar field $f$ over $S = r(T)$ is

\iint_S f\,dS = \iint_T f(r(u, v))\,\lVert r_u\times r_v\rVert\,du\,dv.

The parametrization must be smooth, so that $\lVert N\rVert > 0$ and $dS$ is a genuine area element, and $f$ must be continuous on a set containing $S$ . With $f = 1$ this is the area; if $f = \rho$ is a mass per unit area, it is the mass of the sheet.

In one lineFor a continuous scalar field $f$ and a smooth parametrization $r: T \to S$ , $\iint_S f\,dS = \iint_T f(r(u, v))\,\lVert r_u\times r_v\rVert\,du\,dv$ .

Why it existsIt adds up $f$ times area over the patches. With $f = 1$ it is the area, with $f$ a density it is the mass.

Watch out· The factor that cannot be dropped

The parameter region has area $\iint_T du\,dv$ and the surface has area $\iint_T\lVert N\rVert\,du\,dv$ . They agree only when $\lVert N\rVert = 1$ . For a scalar integral you need $\lVert N\rVert$ , once, as a factor. Never replace $dS$ by $du\,dv$ .

Worked example MATH 376 worksheet 13 problems 3 and 4

Compute

\iint_S z^2\,dS

over the cylinder

x^2 + y^2 = 9

0\le z\le3

, and

\iint_S(x + y + z)\,dS

over the top face

z = 1

of the unit cube.

Solution.

Cylinder, parametrize. $r(\theta, z) = (3\cos\theta,\ 3\sin\theta,\ z)$ , $0\le\theta\le2\pi$ , $0\le z\le3$ . Then $r_\theta\times r_z = (3\cos\theta,\ 3\sin\theta,\ 0)$ , of length $3$ .
Cylinder, integrate. $\int_0^{2\pi}\!\int_0^3z^2\cdot3\,dz\,d\theta = 3\cdot2\pi\cdot9 = 54\pi$ . The integrand does not depend on $\theta$ . $\int_0^3z^2\,dz = 9$ .
Top face. $r(x, y) = (x, y, 1)$ has $N = (0, 0, 1)$ and $\lVert N\rVert = 1$ , so $dS = dA$ . A flat horizontal face is not stretched.
Top face, integrate. $\int_0^1\!\int_0^1(x + y + 1)\,dx\,dy = \int_0^1(\tfrac12 + y + 1)\,dy = \tfrac12 + \tfrac12 + 1 = 2$ .

Answer. $54\pi$ and $2$ .

Worked example MATH 376 homework 5 problem 2

Compute

\iint_S(x^2 + y^2 + z^2)\,dS

over the triangle with vertices

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

Solution.

Plane and graph. The triangle lies in $x + y + z = 1$ , so $z = 1 - x - y$ over $D = \{x, y\ge0,\ x + y\le1\}$ .
Normal. $N = (-g_x, -g_y, 1) = (1, 1, 1)$ , so $\lVert N\rVert = \sqrt3$ , a constant, and $dS = \sqrt3\,dA$ .
Use symmetry. Swapping the names of two coordinates maps the triangle onto itself and keeps $dS$ , so $\iint x^2\,dS = \iint y^2\,dS = \iint z^2\,dS$ . This turns a three-term integral into three times one.
Integrate one term. $3\sqrt3\iint_Dx^2\,dA = 3\sqrt3\int_0^1x^2(1 - x)\,dx = 3\sqrt3\big(\tfrac13 - \tfrac14\big) = 3\sqrt3\cdot\tfrac1{12} = \tfrac{\sqrt3}4$ .

Answer. $\dfrac{\sqrt3}4$ .

Worked example MATH 376 worksheet 12 problem 8

Find the mass of the surface

z = x^2 + y^2

x^2 + y^2\le1

, with density

\rho = 2 + z

Solution.

Normal. $N = (-2x, -2y, 1)$ , $\lVert N\rVert = \sqrt{1 + 4r^2}$ in polar coordinates, and $\rho = 2 + r^2$ .
Set up. $M = \int_0^{2\pi}\!\int_0^1(2 + r^2)\sqrt{1 + 4r^2}\,r\,dr\,d\theta$ . Here $dA = r\,dr\,d\theta$ is the Jacobian of polar coordinates (6.4), separate from the $\lVert N\rVert$ factor.
Substitute. $s = 1 + 4r^2$ , $r\,dr = ds/8$ , $r^2 = (s-1)/4$ , $2 + r^2 = (s + 7)/4$ , limits $s = 1$ to $5$ . Then $M = 2\pi\cdot\frac14\cdot\frac18\int_1^5(s + 7)s^{1/2}\,ds = \frac\pi{16}\int_1^5(s^{3/2} + 7s^{1/2})\,ds$ .
Integrate. $\frac\pi{16}\big[\frac25s^{5/2} + \frac{14}3s^{3/2}\big]_1^5 = \frac\pi{16}\Big(\frac{100\sqrt5}{3} - \frac{76}{15}\Big)$ . At $s = 5$ : $\frac25\cdot25\sqrt5 + \frac{14}3\cdot5\sqrt5 = 10\sqrt5 + \frac{70\sqrt5}3$ . At $s = 1$ : $\frac25 + \frac{14}3 = \frac{76}{15}$ .
Simplify. $\frac\pi{16}\cdot\frac{500\sqrt5 - 76}{15} = \dfrac{\pi(125\sqrt5 - 19)}{60}\approx13.64$ .

Answer. $\dfrac{\pi(125\sqrt5 - 19)}{60}\approx13.64$ . The same method gives $\iint_S z\,dS = \frac{\pi(175\sqrt5 - 41)}{60}\approx18.34$ for $z = 4 - x^2 - y^2$ over $r\le1$ (worksheet 27 problem 8; the course solution stops at the antiderivative, and this value was recomputed).

Your answerCompute

\iint_S z\,dS

over the upper unit hemisphere

x^2 + y^2 + z^2 = 1

z\ge0

Orientation and flux

For a flow we need more than area. A surface has two sides, and flux counts a crossing as positive when it goes one way and negative when it goes the other. We must say which way is positive.

Definition T

Oriented surface

An orientation of a surface $S$ is a continuous choice of unit normal $n$ at every point of $S$ . A surface with an orientation is an oriented surface. A given smooth surface that has one orientation has exactly two, $n$ and $-n$ . A problem tells you which: ‘upward’ for a graph, ‘outward’ for a closed surface such as a sphere, ‘downward’, ‘inward’, or ‘in the direction of $r_u\times r_v$ ’.

In one lineA surface together with a continuous choice of unit normal vector $n$ at every point.

Why it existsFlux changes sign if the normal is reversed, so a flux problem is meaningless until a normal has been chosen: upward, outward, downward, inward.

Definition T

Orientable surface

A surface is orientable if a continuous unit normal field exists on it. A Moebius strip (a band of paper with one half twist, glued) is not: carry a normal vector once around the band and it comes back reversed, so no continuous choice exists. In the picture at the top of the chapter, choose the Moebius strip and compare $N$ at $u = 0$ and $u = 2\pi$ . A parametrization such as $r(u, v) = \big((1 + v\cos\tfrac u2)\cos u,\ (1 + v\cos\tfrac u2)\sin u,\ v\sin\tfrac u2\big)$ , $0\le u\le2\pi$ , $-\frac12\le v\le\frac12$ traces it, and $r_u\times r_v$ at $u = 0$ is the negative of $r_u\times r_v$ at $u = 2\pi$ at the same point.

In one lineA surface that admits an orientation. Spheres, cylinders, graphs, tori and flat pieces are orientable; the Moebius strip is not.

Why it existsFlux and Stokes' theorem both need a normal, so they only make sense on orientable surfaces.

Definition T

Flux

The flux of a continuous vector field $F$ through the oriented surface $S$ with unit normal $n$ is

\iint_S F\cdot n\,dS.

Only the component of $F$ along $n$ crosses the surface, so the integrand $F\cdot n$ is the flux density: the rate of crossing per unit area. If $F$ is the velocity of a fluid, the flux is the volume of fluid crossing the surface per unit time, positive in the direction of $n$ .

In one lineThe flux of a vector field $F$ through an oriented surface $S$ with unit normal $n$ is $\iint_S F\cdot n\,dS$ .

Why it existsIt measures how much of the field crosses the surface, counted positive in the direction of $n$ and negative against it.

Why the dot product with $n$ and not something else? Try a flat square of area $A$ in the plane $x = 0$ , with normal $n = (1, 0, 0)$ . If $F = (1, 0, 0)$ the field goes straight through, and in one second a column of fluid of volume $1\cdot A$ crosses. If $F = (0, 2, 0)$ the field slides along the square and nothing crosses. For a constant $F = v$ the volume crossing in one second is the volume of the slanted column on the square with edge $v$ , which is $(v\cdot n)\,A$ .

Theorem T MATH 376 lectures of 25 and 27 February; Apostol 12.7 and 12.9

Flux as a double integral

Let $r: T\to S$ be a smooth parametrization and $F$ continuous on $S$ . If $N = r_u\times r_v$ points in the direction of the required normal $n$ , then $\iint_SF\cdot n\,dS = \iint_TF(r(u, v))\cdot N(u, v)\,du\,dv$ . If $N$ points the other way, then $n = -N/\lVert N\rVert$ and the flux is $-\iint_TF(r)\cdot N\,du\,dv$ .

In one lineIf $r_u\times r_v$ points the same way as $n$ , then $\iint_S F\cdot n\,dS = \iint_TF(r(u, v))\cdot(r_u\times r_v)\,du\,dv$ ; if it points the opposite way, the right side has the opposite sign.

Why it is trueThe length of the normal cancels, so a flux integral never contains a square root. This one identity is the working formula for every flux problem.

ProofMATH 376 lecture of 27 February

The idea: The unit normal divides by the length of N and the area element multiplies by the length of N. They cancel.

\iint_SF\cdot n\,dS = \iint_TF(r)\cdot n\,\lVert N\rVert\,du\,dv

Surface integral formula with

f = F\cdot n

n\,\lVert N\rVert = \dfrac{N}{\lVert N\rVert}\lVert N\rVert = N

N

points the way

n

does, then

n = N/\lVert N\rVert

at every point.This is where the direction check matters. For 'upward', look at the sign of the third component of

N

. For 'outward' from a sphere or cylinder, test

N

against the position vector at one point.

= \iint_TF(r)\cdot N\,du\,dv

Substitute. The normalization has vanished.

The integral is

-\iint_TF(r)\cdot N\,du\,dv

\blacksquare

If instead

N

points against

n

, then

n = -N/\lVert N\rVert

and the same cancellation gives

n\,\lVert N\rVert = -N

.Do not flip the sign by taking an absolute value at the end: a flux can truly be negative.

The method for flux: (1) parametrize; (2) compute $N = r_u\times r_v$ ; (3) compare its direction with the required orientation, and replace $N$ by $-N$ if they disagree; (4) substitute the surface into $F$ ; (5) dot with $N$ ; (6) integrate over the parameter region. Nothing is normalized.

Why· Another way to write the same flux: Apostol 12.9

Apostol writes the flux of $F = (P, Q, R)$ as $\iint_SP\,dy\wedge dz + Q\,dz\wedge dx + R\,dx\wedge dy$ . The symbol $\iint_SP\,dy\wedge dz$ means $\iint_TP(r(u, v))\,\dfrac{\partial(y, z)}{\partial(u, v)}\,du\,dv$ , where $\dfrac{\partial(y, z)}{\partial(u, v)} = y_uz_v - y_vz_u$ is a $2\times2$ determinant. These three determinants are exactly the three components of $r_u\times r_v$ : the first component of $N$ is $y_uz_v - z_uy_v$ , the second is $z_ux_v - x_uz_v$ and the third is $x_uy_v - y_ux_v$ . So the notation says nothing new. It says that $P$ is paired with the shadow of the patch on the $yz$ -plane, $Q$ with its shadow on the $zx$ -plane and $R$ with its shadow on the $xy$ -plane. 6.7 explains what $dy\wedge dz$ is.

Why· The picture behind the cancellation: a slanted box

During a short time $\Delta t$ , the fluid that crosses a small patch is the fluid in a slanted box. The box has sides $F\Delta t$ (how far the fluid moves), $r_u\Delta u$ and $r_v\Delta v$ (the patch). The volume of a box is the triple product $F\Delta t\cdot(r_u\Delta u\times r_v\Delta v)$ : base area times height, where the height is the component of $F\Delta t$ along the normal. Divide by $\Delta t$ and add the boxes: $\iint F\cdot(r_u\times r_v)\,du\,dv$ , with no square root anywhere. The length of $N$ is the base area and the unit normal only extracts the height.

Worked example MATH 376 Midterm 2 problem 1(b), 4 points; see [A.3](/vol2/a3-midterm-2/)

Find the flux of

F = (x, y, z)

through

S = \{(u, v, e^u) : 0\le u, v\le1\}

with the normal pointing in the positive

z

direction.

Solution.

Normal. $N = r_u\times r_v = (-e^u, 0, 1)$ , with positive third component. From the previous example. It already points the required way.
Substitute. $F(r(u, v)) = (u, v, e^u)$ . $x = u$ , $y = v$ , $z = e^u$ .
Dot. $F\cdot N = -ue^u + 0 + e^u = (1 - u)e^u$ . No square root: $n\,dS = N\,du\,dv$ .
Set up. $\int_0^1\!\int_0^1(1 - u)e^u\,dv\,du = \int_0^1(1 - u)e^u\,du$ . The integrand has no $v$ , so the $v$ integral over $[0,1]$ is $1$ .
Antiderivative. Integration by parts with $g = 1 - u$ and $dh = e^u\,du$ gives $\int(1 - u)e^u\,du = (1 - u)e^u + \int e^u\,du = (2 - u)e^u$ . Check by differentiating: $\frac d{du}(2 - u)e^u = -e^u + (2 - u)e^u = (1 - u)e^u$ .
Evaluate. $\big[(2 - u)e^u\big]_0^1 = 1\cdot e - 2\cdot1 = e - 2$ .

Answer. $e - 2\approx0.718$ . The practice exam version has $S = \{(u, v, v - u^3) : (u, v)\in[0,1]\times[0,2]\}$ and $F = (x, 2y, z)$ . There $N = (3u^2, -1, 1)$ (positive third component again), $F(r)\cdot N = 3u^3 - 2v + v - u^3 = 2u^3 - v$ , and $\int_0^1\!\int_0^2(2u^3 - v)\,dv\,du = \int_0^1(4u^3 - 2)\,du = 1 - 2 = -1$ .

Worked example MATH 376 lecture of 25 February, Example 5.5

Find the flux of

F = (x, 0, 2z)

through the paraboloid patch

S = \{(u, v, u^2 + v^2) : (u, v)\in[0, 1]^2\}

with the orientation of

r_u\times r_v

Solution.

Tangent vectors and cross. $r_u = (1, 0, 2u)$ , $r_v = (0, 1, 2v)$ , $N = (-2u, -2v, 1)$ . Graph formula with $g = u^2 + v^2$ .
Substitute and dot. $F(r) = (u, 0, 2(u^2 + v^2))$ and $F\cdot N = -2u^2 + 0 + 2u^2 + 2v^2 = 2v^2$ . The $u^2$ terms cancel.
Integrate. $\int_0^1\!\int_0^12v^2\,du\,dv = \int_0^12v^2\,dv = \frac23$ .

Answer. $\dfrac23$ .

Worked example MATH 376 worksheet 11 problem 6; worksheet 12 problem 5 is the field $(0, 0, z)$

Find the upward flux of

F = (y, -x, z)

through the cap

z = 4 - x^2 - y^2

z\ge0

, using polar coordinates.

Solution.

Parametrize. $r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ 4 - r^2)$ , $0\le r\le2$ , $0\le\theta\le2\pi$ .
Cross. $N = r_r\times r_\theta = (2r^2\cos\theta,\ 2r^2\sin\theta,\ r)$ . The same computation as in the area example: the constant in $z$ disappears on differentiating. The third component $r\ge0$ , so it points up.
Substitute. $F(r) = (r\sin\theta,\ -r\cos\theta,\ 4 - r^2)$ .
Dot. $F\cdot N = 2r^3\sin\theta\cos\theta - 2r^3\cos\theta\sin\theta + (4 - r^2)r = 4r - r^3$ . The field $(y, -x, 0)$ circles the axis and is tangent to every horizontal circle, so it contributes nothing.
Integrate. $\int_0^{2\pi}\!\int_0^2(4r - r^3)\,dr\,d\theta = 2\pi\big[2r^2 - \tfrac{r^4}4\big]_0^2 = 2\pi(8 - 4) = 8\pi$ .

Answer. $8\pi$ . For $F = (0, 0, z)$ the graph shortcut $n\,dS = (2x, 2y, 1)\,dA$ gives the integrand $4 - x^2 - y^2$ and the same $8\pi$ .

Worked example MATH 376 worksheet 12 problems 6 and 7

Find the outward flux of

F = (x, y, 0)

through the lateral surface of the cylinder

x^2 + y^2 = 1

0\le z\le2

, and the outward flux of

F = (x, y, z)

through the upper unit hemisphere by a shortcut.

Solution.

Cylinder. $r(\theta, z) = (\cos\theta,\ \sin\theta,\ z)$ gives $N = r_\theta\times r_z = (\cos\theta,\ \sin\theta,\ 0)$ , which points away from the axis: outward. Test at $\theta = 0$ : the point is $(1, 0, z)$ and $N = (1, 0, 0)$ .
Cylinder, dot. $F(r) = (\cos\theta, \sin\theta, 0)$ , so $F\cdot N = \cos^2\theta + \sin^2\theta = 1$ , and the flux is $\int_0^{2\pi}\!\int_0^21\,dz\,d\theta = 4\pi$ .
Hemisphere, shortcut. On the unit sphere the outward unit normal at $(x, y, z)$ is $(x, y, z)$ itself, so $F\cdot n = x^2 + y^2 + z^2 = 1$ , constant. If $F\cdot n$ is constant $c$ on $S$ , the flux is $c\cdot\operatorname{area}(S)$ .
Area. Half of $4\pi$ , so the flux is $1\cdot2\pi = 2\pi$ .

Answer. $4\pi$ (check: $F\cdot n = 1$ times the area $2\pi\cdot1\cdot2 = 4\pi$ ) and $2\pi$ . Two more shortcuts in the same worksheets: on a flat horizontal piece with $n = (0, 0, 1)$ only the third component of $F$ matters, since the first two are tangent to the piece. So the flux of $(0, 0, 5)$ upward through the disc $x^2 + y^2\le16$ in $z = 2$ is $5\cdot16\pi = 80\pi$ , and that of $(x, y, 5)$ through $x^2 + y^2\le4$ in $z = 3$ is $5\cdot4\pi = 20\pi$ .

Worked example MATH 376 worksheet 11 problem 7; worksheet 13 problem 6 is $F = (x, y, 0)$, $0\le z\le4$, oriented upward and inward

Find the flux of

F = (0, 0, z^2)

through the cone

z = \sqrt{x^2 + y^2}

1\le z\le2

, oriented downward.

Solution.

Parametrize. $r(r, \theta) = (r\cos\theta,\ r\sin\theta,\ r)$ with $1\le r\le2$ . On the cone $z = r$ .
Cross. $r_r\times r_\theta = (-r\cos\theta,\ -r\sin\theta,\ r)$ . Its third component is $r>0$ : it points up (and in). The problem wants downward: the wrong way.
Flip. Use $N = -r_r\times r_\theta = (r\cos\theta,\ r\sin\theta,\ -r)$ . By the theorem, the flux is the integral with $-N$ .
Dot. $F\cdot N = z^2\cdot(-r) = r^2(-r) = -r^3$ . $z = r$ on the cone.
Integrate. $\int_0^{2\pi}\!\int_1^2-r^3\,dr\,d\theta = 2\pi\big[-\tfrac{r^4}4\big]_1^2 = 2\pi\big(-4 + \tfrac14\big) = -\dfrac{15\pi}2$ .

Answer. $-\frac{15\pi}2$ . It is negative because the field points up and the surface is oriented down. For the worksheet 13 version the vector $r_r\times r_\theta$ already points up and inward, so there is no flip: $F\cdot N = (r\cos\theta)(-r\cos\theta) + (r\sin\theta)(-r\sin\theta) = -r^2$ , and the flux is $2\pi\big[-\frac{r^3}3\big]_0^4 = -\frac{128\pi}3$ .

Worked example MATH 376 homework 5 problem 1

Parametrize the ellipsoid

\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1

, find its normal field, and decide whether it points in or out.

Solution.

Parametrize. $r(u, v) = (a\sin v\cos u,\ b\sin v\sin u,\ c\cos v)$ , $0\le u\le2\pi$ , $0\le v\le\pi$ . The sphere parametrization with each coordinate stretched. It satisfies the equation: $\sin^2v(\cos^2u + \sin^2u) + \cos^2v = 1$ .
Tangent vectors. $r_u = (-a\sin v\sin u,\ b\sin v\cos u,\ 0)$ and $r_v = (a\cos v\cos u,\ b\cos v\sin u,\ -c\sin v)$ .
Cross. $N = (-bc\sin^2v\cos u,\ -ac\sin^2v\sin u,\ -ab\sin v\cos v)$ .
Test the direction. At $u = 0$ , $v = \pi/2$ the point is $(a, 0, 0)$ and $N = (-bc, 0, 0)$ , which points toward the origin. One test point is enough for a continuous normal that never vanishes inside the region.

Answer. The normal $r_u\times r_v$ points inward. For a flux problem that says ‘outward’, use $-N = (bc\sin^2v\cos u,\ ac\sin^2v\sin u,\ ab\sin v\cos v)$ .

Worked example MATH 376 practice final problem 2; see [A.5](/vol2/a5-practice-exams/)

Find the upward flux of

f = (y, x, 1)

through

S = \{(x, y, x^2 - y^2) : x, y\in[0, 1]\}

Solution.

Cross. $r_x = (1, 0, 2x)$ , $r_y = (0, 1, -2y)$ , so $N = (-2x,\ 2y,\ 1)$ . Graph formula: $-g_x = -2x$ , $-g_y = 2y$ .
Direction. The third component is $1>0$ : upward, as required.
Dot. $f(r)\cdot N = y(-2x) + x(2y) + 1 = 1$ . $f$ does not depend on $z$ . The $xy$ terms cancel.
Integrate. $\int_0^1\!\int_0^1 1\,dy\,dx = 1$ .

Answer. $1$ .

Your answerFind the outward flux of

F = (x, y, z)

through the lateral surface of the cylinder

x^2 + y^2 = 4

0\le z\le3

Quick checkYou parametrize a surface and

r_u\times r_v

points downward at a point, but the problem asks for the upward flux. What do you do?

Quick checkIn a flux integral through a surface parametrized by

r(u, v)

, what happens to

\lVert r_u\times r_v\rVert

Quick checkThe Moebius strip is not orientable. Which statement about it is correct?

The integral does not depend on the parametrization

The same surface has many parametrizations. The answer must not depend on the one you picked, or the computation would not measure the surface.

Proposition T MATH 376 lecture of 27 February, Proposition 5.11 and Lemma 5.12; Apostol 12.8

Surface integrals do not depend on the parametrization

Let $r: T\to S$ be smooth and $g = (u, v): \tau\to T$ a one-to-one continuously differentiable map with $\det Jg\neq0$ , so that $\rho = r\circ g$ is another parametrization of $S$ with parameters $(s, t)$ . Here $Jg$ is the Jacobian matrix with entries $u_s, u_t, v_s, v_t$ (5.3).

In one lineIf $\rho = r\circ g$ with $g$ one-to-one and continuously differentiable, then $\rho_s\times\rho_t = (\det Jg)\,(r_u\times r_v)\circ g$ ; so a scalar surface integral is the same for $r$ and $\rho$ , and the flux is the same if $\det Jg>0$ and changes sign if $\det Jg<0$ .

Why it is trueIt justifies computing a surface integral with whichever parametrization is easiest, and it shows that orientation is the one thing that depends on the choice.

ProofMATH 376 lecture of 27 February

The idea: The chain rule says each new tangent vector is a combination of the old ones. Crossing two such combinations leaves exactly the determinant of the combination coefficients times the old cross product. Then the change of variables formula (6.4) absorbs the determinant into the area element.

\rho_s = u_s\,r_u + v_s\,r_v

and

\rho_t = u_t\,r_u + v_t\,r_v

Chain rule for

\rho(s, t) = r(u(s, t), v(s, t))

;

r_u, r_v

are evaluated at

g(s, t)

\rho_s\times\rho_t = u_su_t\,r_u\times r_u + u_sv_t\,r_u\times r_v + v_su_t\,r_v\times r_u + v_sv_t\,r_v\times r_v

Cross product is linear in each slot, so multiply out the four terms.

\rho_s\times\rho_t = (u_sv_t - v_su_t)\,r_u\times r_v

A vector crossed with itself is

0

(zero area), and

r_v\times r_u = -r_u\times r_v

(anti-symmetry).The lecture notes print a slip on this line; the factor is

u_sv_t - v_su_t

, which is what the four-term expansion gives.

u_sv_t - v_su_t = \det Jg

, so

\rho_s\times\rho_t = (\det Jg)\,(r_u\times r_v)\circ g

Definition of the determinant of the Jacobian matrix of

g

\lVert\rho_s\times\rho_t\rVert = \lvert\det Jg\rvert\,\lVert(r_u\times r_v)\circ g\rVert

Take lengths. The absolute value appears because a length is not negative.

\displaystyle\iint_Tf(r)\lVert r_u\times r_v\rVert\,du\,dv = \iint_\tau f(\rho)\,\lvert\det Jg\rvert\,\lVert(r_u\times r_v)\circ g\rVert\,ds\,dt = \iint_\tau f(\rho)\lVert\rho_s\times\rho_t\rVert\,ds\,dt

Change of variables (6.4) for the continuous function

f(r(u,v))\lVert r_u \times r_v\rVert

and the map

g

, with the factor

\lvert\det Jg\rvert

The scalar integrals are equal, and the flux integrals are equal or opposite according to the sign of

\det Jg

\blacksquare

For flux, the same chain gives

\rho_s\times\rho_t

equal to

\det Jg

times the old vector. If

\det Jg>0

it points the same way and the flux is unchanged; if

\det Jg<0

it points the opposite way and the flux changes sign.So a surface integral depends on the surface and on the orientation, not on the parametrization. Swapping the names of

u

and

v

has

\det Jg = -1

and reverses the orientation.

Proof workshopGive the reason for each line: the new normal is the Jacobian determinant times the old oneMATH 376 lecture of 27 February

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\rho_s = u_s\,r_u + v_s\,r_v$ and $\rho_t = u_t\,r_u + v_t\,r_v$ .
$\rho_s\times\rho_t = u_su_t\,r_u\times r_u + u_sv_t\,r_u\times r_v + v_su_t\,r_v\times r_u + v_sv_t\,r_v\times r_v$ .
$r_u\times r_u = 0$ and $r_v\times r_v = 0$ .
$r_v\times r_u = -\,r_u\times r_v$ .
$\rho_s\times\rho_t = (u_sv_t - v_su_t)\,r_u\times r_v$ .
$u_sv_t - v_su_t = \det Jg$ .

Where it is used

Used inside maths· Inside maths: Archimedes, mean curvature and the divergence theorem

The area of the sphere $4\pi R^2$ is on the one hand a computation with $r_\varphi\times r_\theta$ , and on the other the number Archimedes found by comparing a sphere with the cylinder around it: the zone between two parallel planes has the same area as the corresponding band of the cylinder, $2\pi Rh$ . The word ‘area’ also leads to curvature: the cross product’s length squared is $EG - F^2$ , with $E = r_u\cdot r_u$ , $F = r_u\cdot r_v$ , $G = r_v\cdot r_v$ , the first fundamental form of the surface, which is where differential geometry begins. And flux is the left side of the divergence theorem in 6.6, where the whole surface integral for a closed surface becomes a volume integral. The flux of $F = (x, y, z)$ out of a torus of radii $R$ and $a$ is $6\pi^2Ra^2$ (dot $F$ with the outward normal $a(R + a\cos v)(\cos u\cos v, \sin u\cos v, \sin v)$ found above and integrate), exactly three times its volume $2\pi^2Ra^2$ . Why a factor of exactly $3$ is the subject of 6.6.

Used in the real world· Used in the real world: a solar panel and a mesh

Sunlight delivers about $1000$ W per square metre on a surface facing the sun; the field is $F = 1000\,d$ , with $d$ the unit vector along the rays, and the power on a panel is the flux through it. A flat roof panel of $1.7$ m $^2$ with upward unit normal $n = (0, 0, 1)$ sees the sun $30^\circ$ from the vertical, so the rays travel along $d = (-\tfrac12, 0, -\tfrac{\sqrt3}2)$ , and the power on the panel is $\lvert F\cdot n\rvert\cdot\text{area} = 1000\cdot\tfrac{\sqrt3}2\cdot1.7\approx1472$ W, about $87\%$ of the full $1700$ W: the tilt costs $\cos30^\circ$ . Tilting the panel to face the sun turns the dot product into $1000$ and the power into $1700$ W.

A 3D model in a game engine or a 3D printer file is a list of flat triangles. The software finds the area of each triangle as half the length of the cross product of two edges, exactly as in the worked example, and adds: a 10 cm cube stored as $12$ triangles has area $12\cdot\frac12\cdot100 = 600$ cm $^2$ , which is what determines the amount of paint or resin for the skin. The unit normal of each triangle, the cross product divided by its length, is what the shading uses: the brightness of a pixel on a triangle is proportional to the dot product of that normal with the direction to the light. A hemispherical dome of radius $12$ m has roof area $2\pi\cdot12^2\approx904.8$ m $^2$ , the number that matters for the cost of the roofing.

Your answerA flat solar panel of

1.7

^2

lies on a horizontal roof. Sunlight of intensity

1000

W/m

^2

arrives along the direction

(-\frac12, 0, -\frac{\sqrt3}2)

. How many watts reach the panel?

Traps

Watch out· Midterm 2 problem 1(b) lost two of four points: three mistakes in one flux integral

On the 376 midterm, a student wrote $dS = du\,dv$ but kept the unit normal $n = N/\sqrt{e^{2u} + 1}$ in the integrand, so the stray square root stayed. The real area element is $dS = \sqrt{e^{2u} + 1}\,du\,dv$ , and it cancels the denominator of $n$ . They then pulled the factor $1/\sqrt{e^{2u} + 1}$ out of the integral over $u$ , which is never allowed for a factor that depends on $u$ . A sign that something went wrong is that their boxed answer still contained the variable $u$ : a definite integral over the whole surface is a number. And they integrated $ue^u$ as $\frac{u^2}2e^u - e^u$ , which is not integration by parts; the antiderivative is $ue^u - e^u$ . Differentiate every antiderivative before they use it. The rubric item applied was ‘correct except for a major computational mistake’. The full solution is in A.3 and A.1.

Watch out· Compare the sign of the cross product with the orientation, every time

The formula $\iint F\cdot n\,dS = \iint F\cdot(r_u\times r_v)\,du\,dv$ holds only when $r_u\times r_v$ points the way $n$ does. Test it at one point: the sign of the third component for upward or downward, the direction relative to the axis or the centre for outward or inward. The cone $(r\cos\theta, r\sin\theta, r)$ gives a vector pointing up and inward, and the ellipsoid parametrization gives a vector pointing inward. Both are traps in the worksheets.

Watch out· Do not normalize, and do not drop the length, whichever integral it is

Scalar integral: $f\,\lVert N\rVert\,du\,dv$ , with the length. Flux: $F\cdot N\,du\,dv$ , with no length and no unit vector. Writing the unit normal and also the length (the other way round from the exam mistake) gives the right answer but is slow and invites arithmetic slips with the square root.

Watch out· Let the problem's constraints set the parameter limits

For the paraboloid cap above the plane $z = 0$ the limit is $r\le\sqrt3/2$ , from $z\ge0$ , not $r\le1$ or $r\le3/4$ . For the cone between $z = 0$ and $z = 2^{1/4}$ the limit is on $r$ , not on $\theta$ . Write the constraint as an inequality before you integrate. A dropped absolute value is the same kind of slip in a different place: the set $\max(\lvert x\rvert, \lvert y\rvert, \lvert z\rvert) = 1$ is the surface of the cube $[-1, 1]^3$ , not of a cube in one octant. 6.6 returns to it.

Watch out· Cone apex, sphere poles

At the apex of the cone and at the poles of the sphere ( $\varphi = 0, \pi$ ) the vector $N$ is $0$ and the parametrization is not regular. These are points or curves that do not carry area, so they change no integral. What you must not do is divide by $\lVert N\rVert$ there.

Problems

checknew (MATH 376 worksheet 12 style)

Compute $(1, 2, -1)\times(3, 0, 2)$ and verify that it is orthogonal to both vectors.

Solution

Components: $2\cdot2 - (-1)\cdot0 = 4$ ; $(-1)\cdot3 - 1\cdot2 = -5$ ; $1\cdot0 - 2\cdot3 = -6$ . So the cross product is $(4, -5, -6)$ . Check: $(4, -5, -6)\cdot(1, 2, -1) = 4 - 10 + 6 = 0$ and $(4, -5, -6)\cdot(3, 0, 2) = 12 + 0 - 12 = 0$ .

checknew (MATH 376 worksheet 11 problem 5 style)

Find $r_u\times r_v$ for $r(u, v) = (u + v,\ u - v,\ uv)$ at $(u, v) = (1, 1)$ .

Solution

$r_u = (1, 1, v) = (1, 1, 1)$ and $r_v = (1, -1, u) = (1, -1, 1)$ at the point. $r_u\times r_v = (1\cdot1 - 1\cdot(-1),\ 1\cdot1 - 1\cdot1,\ 1\cdot(-1) - 1\cdot1) = (2, 0, -2)$ . Check: dotted with $r_u$ gives $2 - 2 = 0$ , with $r_v$ gives $2 - 2 = 0$ .

checknew (Midterm 2 problem 1(b) mistake)

A cylinder of radius $4$ and height $5$ is parametrized on $[0, 2\pi]\times[0, 5]$ . What is the area of the parameter rectangle, and what is the area of the cylinder? What is $\lVert r_\theta\times r_z\rVert$ ?

Solution

The rectangle has area $10\pi$ . $r = (4\cos\theta, 4\sin\theta, z)$ has $r_\theta\times r_z = (4\cos\theta, 4\sin\theta, 0)$ , of length $4$ . So the cylinder has area $4\cdot10\pi = 40\pi$ , which is $2\pi\cdot4\cdot5$ , circumference times height.

check

True or false: if you reparametrize a surface by swapping the roles of $u$ and $v$ , the flux of $F$ through it stays the same.

Solution

It depends on what is held fixed. Swapping $u$ and $v$ replaces $r_u\times r_v$ by $r_v\times r_u = -r_u\times r_v$ , so if you keep using $\iint F\cdot(r_u\times r_v)\,du\,dv$ without comparing the direction with the required orientation, the sign of the answer flips: false. If you do compare, and use $-N$ when $N$ points the wrong way, the flux through the oriented surface does not depend on the parametrization: true. What changes is which of $\pm N$ equals $n$ .

usenew (MATH 376 worksheet 12 problem 3 style)

Find the area of the part of the plane $x + 2y + 3z = 6$ in the first octant.

Solution

The plane meets the axes at $A = (6, 0, 0)$ , $B = (0, 3, 0)$ , $C = (0, 0, 2)$ . $\overrightarrow{AB} = (-6, 3, 0)$ and $\overrightarrow{AC} = (-6, 0, 2)$ , so $\overrightarrow{AB}\times\overrightarrow{AC} = (3\cdot2 - 0,\ 0\cdot(-6) - (-6)\cdot2,\ (-6)\cdot0 - 3\cdot(-6)) = (6, 12, 18)$ , of length $6\sqrt{14}$ . The area is $3\sqrt{14}\approx11.22$ . Check by the graph formula: $z = (6 - x - 2y)/3$ , $\lVert N\rVert = \sqrt{1 + \frac19 + \frac49} = \frac{\sqrt{14}}3$ , and the region $D$ under it has area $\frac12\cdot6\cdot3 = 9$ , giving $3\sqrt{14}$ .

usenew (symmetry, compare MATH 376 homework 5 problem 2)

Compute $\iint_Sx^2\,dS$ over the unit sphere.

Solution

By symmetry (swapping coordinate names maps the sphere to itself), $\iint x^2\,dS = \iint y^2\,dS = \iint z^2\,dS$ . Their sum is $\iint(x^2 + y^2 + z^2)\,dS = \iint1\,dS = 4\pi$ . So each is $\frac{4\pi}3$ .

usenew (compare MATH 376 worksheet 12 problem 7)

Find the outward flux of the constant field $F = (1, 2, 3)$ through the upper hemisphere of radius $2$ .

Solution

Use $r(\varphi, \theta) = 2(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$ , $0\le\varphi\le\pi/2$ . $N = r_\varphi\times r_\theta = 4\sin\varphi\,(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$ , which points outward. $F\cdot N = 4\sin\varphi\,(\sin\varphi\cos\theta + 2\sin\varphi\sin\theta + 3\cos\varphi)$ . The $\theta$ integral of the first two terms over $[0, 2\pi]$ is $0$ . What is left is $12\sin\varphi\cos\varphi$ , and $2\pi\int_0^{\pi/2}12\sin\varphi\cos\varphi\,d\varphi = 2\pi\cdot6 = 12\pi$ . This is $3\cdot\pi\cdot2^2$ , the third component times the area of the disc under the hemisphere, as it must be: only the vertical part of a constant field sees the shadow.

useMATH 376 homework 5 problem 3

Find the outward flux of $F = (x^2, y^2, z^2)$ through the cylinder $x^2 + y^2 = 1$ , $-1\le z\le1$ (lateral surface only).

Solution

$r = (\cos\theta, \sin\theta, z)$ , $N = (\cos\theta, \sin\theta, 0)$ (outward). $F\cdot N = \cos^3\theta + \sin^3\theta$ , with no $z$ and no third-component term. $\int_{-1}^1\int_0^{2\pi}(\cos^3\theta + \sin^3\theta)\,d\theta\,dz = 2\big(\int_0^{2\pi}(1 - \sin^2\theta)\cos\theta\,d\theta + \int_0^{2\pi}(1 - \cos^2\theta)\sin\theta\,d\theta\big)$ . The first is $\big[\sin\theta - \frac{\sin^3\theta}3\big]_0^{2\pi} = 0$ and the second is $\big[-\cos\theta + \frac{\cos^3\theta}3\big]_0^{2\pi} = 0$ . The flux is $0$ .

useMATH 376 worksheet 12 problem 4

Find the upward flux of $F = (x, 2y, z)$ through the part of the plane $z = 4 - x - 2y$ over $0\le x\le2$ , $0\le y\le1$ .

Solution

$r(x, y) = (x, y, 4 - x - 2y)$ , $N = (-g_x, -g_y, 1) = (1, 2, 1)$ , third component $+1$ so upward as required. $F(r) = (x, 2y, 4 - x - 2y)$ , so $F\cdot N = x + 4y + 4 - x - 2y = 4 + 2y$ . The $x$ terms cancel. $\int_0^1\!\int_0^2(4 + 2y)\,dx\,dy = \int_0^1(8 + 4y)\,dy = 8 + 2 = 10$ .

useMATH 376 worksheet 27 problem 9

Find the upward flux of $F = (x, y, z)$ through $z = 4 - x^2 - y^2$ over the disc $x^2 + y^2\le1$ .

Solution

$N = (2x, 2y, 1)$ (upward), $F\cdot N = 2x^2 + 2y^2 + (4 - x^2 - y^2) = 4 + x^2 + y^2$ . In polar coordinates $\int_0^{2\pi}\!\int_0^1(4 + r^2)\,r\,dr\,d\theta = 2\pi\big[2r^2 + \tfrac{r^4}4\big]_0^1 = 2\pi\cdot\tfrac94 = \tfrac{9\pi}2$ . Compare the cap over $r\le2$ , where the same field gives $24\pi$ in the picture at the top of the chapter.

useMATH 376 worksheet 14 problem 3

The field $B = \dfrac{(-y, x, 0)}{x^2 + y^2}$ is defined away from the $z$ -axis. Find its upward flux through the part of the plane $z = y + 5$ inside the cylinder $x^2 + y^2 = 1$ .

Solution

Graph of $g = y + 5$ over the unit disc, so $N = (-g_x, -g_y, 1) = (0, -1, 1)$ , upward. $B\cdot N = \dfrac{-x}{x^2 + y^2}$ . In polar coordinates $\int_0^{2\pi}\!\int_0^1\dfrac{-r\cos\theta}{r^2}\,r\,dr\,d\theta = \int_0^{2\pi}\!\int_0^1(-\cos\theta)\,dr\,d\theta = \int_0^{2\pi}(-\cos\theta)\,d\theta = 0$ . The $r$ from the area element cancels one power of $r^2$ . The plane passes through the $z$ -axis at $(0, 0, 5)$ , where $B$ is not defined. The integrand times $r$ is bounded, so the integral exists, but the field is undefined at that one point.

useMATH 376 worksheet 14 problem 4

A bowl is the part of the sphere $x^2 + y^2 + z^2 = 4$ with $z\le0$ . Fluid of pressure $10 - z$ pushes on it. Find the vertical force $F_z = \iint_S(10 - z)(-n_z)\,dS$ , where $n$ is the outward normal.

Solution

On the sphere of radius $2$ , $n = (x, y, z)/2$ , so $n_z = z/2 = \cos\varphi$ , and $z = 2\cos\varphi$ . By the sphere computation, $dS = 4\sin\varphi\,d\varphi\,d\theta$ . The lower half is $\pi/2\le\varphi\le\pi$ . The integrand is $(10 - 2\cos\varphi)(-\cos\varphi) = -10\cos\varphi + 2\cos^2\varphi$ . So $F_z = \int_0^{2\pi}\int_{\pi/2}^{\pi}(-10\cos\varphi + 2\cos^2\varphi)\,4\sin\varphi\,d\varphi\,d\theta$ . Put $c = \cos\varphi$ , $dc = -\sin\varphi\,d\varphi$ , limits $c = 0$ to $-1$ , which reverses to $\int_{-1}^0(-10c + 2c^2)\,dc = \big[-5c^2 + \frac23c^3\big]_{-1}^0 = 0 - (-5 - \frac23) = \frac{17}3$ . So $F_z = 2\pi\cdot4\cdot\frac{17}3 = \frac{136\pi}3$ . (The course solution as extracted shows $\frac{136}3$ with the $\pi$ lost; $\frac{136\pi}3$ is correct.)

provenew (the triple product and flux of a constant field)

Let $a, b$ be vectors in space and $P = \{ua + vb : 0\le u, v\le1\}$ the parallelogram with normal $n = (a\times b)/\lVert a\times b\rVert$ . Show that the flux of a constant field $c$ through $P$ is $c\cdot(a\times b)$ .

Solution

Parametrize $r(u, v) = ua + vb$ on $[0, 1]^2$ . Then $r_u = a$ , $r_v = b$ , $N = a\times b$ is constant and points the way $n$ does. By the flux theorem, $\iint_PF\cdot n\,dS = \iint_{[0,1]^2}c\cdot(a\times b)\,du\,dv = c\cdot(a\times b)\cdot1$ . This is the volume of the parallelepiped on $a, b, c$ (up to sign), which is the slanted-column picture.

provenew (Archimedes' hat-box, Apostol 12.5)

Prove that the area of the zone of the sphere of radius $R$ between the planes $z = z_1$ and $z = z_2$ (with $-R\le z_1<z_2\le R$ ) is $2\pi R(z_2 - z_1)$ .

Solution

Use $r(\varphi, \theta)$ of radius $R$ . $\lVert N\rVert = R^2\sin\varphi$ for $0\le\varphi\le\pi$ , as in the example. The zone is $\varphi_1\le\varphi\le\varphi_2$ with $z = R\cos\varphi$ , so $\varphi_1<\varphi_2$ correspond to $z_2>z_1$ . The area is $2\pi\int_{\varphi_1}^{\varphi_2}R^2\sin\varphi\,d\varphi = 2\pi R^2(\cos\varphi_1 - \cos\varphi_2) = 2\pi R(z_2 - z_1)$ since $R\cos\varphi_1 = z_2$ and $R\cos\varphi_2 = z_1$ . The area depends only on the thickness.

provenew (graphs, Apostol 12.5)

For a graph $z = g(x, y)$ over a region $D$ , prove that $\operatorname{area}(S) = \iint_D\sqrt{1 + g_x^2 + g_y^2}\,dA$ , and that if $g$ is linear, $z = px + qy + c$ , then $\operatorname{area}(S) = \sqrt{1 + p^2 + q^2}\operatorname{area}(D)$ .

Solution

The graph formula gives $N = (-g_x, -g_y, 1)$ and $\lVert N\rVert = \sqrt{1 + g_x^2 + g_y^2}$ , so the definition of area gives the first statement. If $g = px + qy + c$ then $g_x = p$ and $g_y = q$ are constants, so $\lVert N\rVert$ is a constant that comes out of the integral, leaving $\operatorname{area}(D)$ .

provenew (orientation reverses under swapping the parameters)

Show that if $\rho(s, t) = r(t, s)$ (swap the parameters), then $\rho_s\times\rho_t = -(r_u\times r_v)$ at corresponding points, and conclude that the flux integral changes sign unless one also flips the required normal.

Solution

$\rho_s = r_v$ and $\rho_t = r_u$ at the point $(t, s)$ , so $\rho_s\times\rho_t = r_v\times r_u = -r_u\times r_v$ by anti-symmetry. This is also the case $g(s, t) = (t, s)$ with $\det Jg = -1$ in the independence theorem. If the problem requires $n$ and the new $\rho_s\times\rho_t$ points against it, the flux is $-\iint F(\rho)\cdot(\rho_s\times\rho_t)\,ds\,dt$ , which equals the flux computed with $r$ . The two parametrizations give the same oriented flux once each is compared with $n$ .

stretchApostol 12.5, the helicoid

Find the area of the helicoid $r(u, v) = (u\cos v,\ u\sin v,\ v)$ for $0\le u\le1$ , $0\le v\le2\pi$ .

Solution

$r_u = (\cos v, \sin v, 0)$ and $r_v = (-u\sin v,\ u\cos v,\ 1)$ , so $r_u\times r_v = (\sin v,\ -\cos v,\ u)$ and $\lVert\cdot\rVert = \sqrt{1 + u^2}$ . The area is $2\pi\int_0^1\sqrt{1 + u^2}\,du = 2\pi\cdot\frac12\big[u\sqrt{1 + u^2} + \ln(u + \sqrt{1 + u^2})\big]_0^1 = \pi\big(\sqrt2 + \ln(1 + \sqrt2)\big)\approx7.21$ .

stretchnew (Schwarz's lantern)

Derive the area of Schwarz’s lantern with $m$ slabs and $n$ points per ring on a cylinder of radius $1$ and height $1$ , and show that for $m = n$ it tends to $2\pi$ as $n\to\infty$ while for $m = n^2$ it does not.

Solution

Neighbouring rings are rotated by $\pi/n$ relative to each other, so each triangle has two corners on one ring, joined by a chord of length $2\sin(\pi/n)$ , and a third corner on the next ring, directly over the middle of the chord in angle. The horizontal distance from that third corner to the chord’s midpoint is $1 - \cos(\pi/n)$ , and the vertical distance is $1/m$ , so the height of the triangle (the distance from the apex to the chord’s midpoint) is $\sqrt{m^{-2} + (1 - \cos(\pi/n))^2}$ . A triangle has area $\sin(\pi/n)\sqrt{m^{-2} + (1 - \cos(\pi/n))^2}$ , and there are $2nm$ of them. So the area is $2nm\sin(\pi/n)\sqrt{m^{-2} + (1 - \cos(\pi/n))^2}$ . Rewrite it as $2n\sin(\pi/n)\sqrt{1 + m^2(1 - \cos(\pi/n))^2}$ . As $n\to\infty$ , $2n\sin(\pi/n)\to2\pi$ and $1 - \cos(\pi/n)\approx\frac{\pi^2}{2n^2}$ . If $m = n$ then $m^2(1 - \cos)^2\approx\frac{\pi^4}{4n^2}\to0$ and the area tends to $2\pi$ . If $m = n^2$ then $m^2(1 - \cos)^2\to\frac{\pi^4}4$ and the area tends to $2\pi\sqrt{1 + \pi^4/4}\approx31.6$ , five times the cylinder’s area. The lab confirms both.

Summary

The cross product $a\times b = (a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1)$ is orthogonal to $a$ and $b$ , and its length is the area of the parallelogram they span. Check every cross product with two dot products.
A parametric surface is $r(u, v)$ on a parameter region. Learn the table: graph, cylinder, cone, paraboloid, sphere, ellipsoid, torus, parallelogram. The problem’s constraints set the parameter limits.
The fundamental vector product $N = r_u\times r_v$ is normal to the surface and $\lVert N\rVert$ is the stretching factor from parameter area to surface area. For a graph $N = (-g_x, -g_y, 1)$ points up.
Area is $\iint\lVert N\rVert\,du\,dv$ ; a scalar integral is $\iint f(r)\lVert N\rVert\,du\,dv$ (mass if $f$ is a density). The sphere has area $4\pi R^2$ , the cone and paraboloid cap are Midterm 2 problems, the torus has $4\pi^2Ra$ .
Flux through an oriented surface is $\iint F\cdot n\,dS = \iint F(r)\cdot N\,du\,dv$ , where $N$ must point the way $n$ does; otherwise use $-N$ . The length of $N$ cancels, so a flux integral has no square root.
Compare the sign of $N$ with ‘upward’, ‘outward’, ‘downward’ or ‘inward’ at one test point, every time. If $F\cdot n$ is constant the flux is that constant times the area.
The integral does not depend on the parametrization except through orientation, because $\rho_s\times\rho_t = (\det Jg)\,r_u\times r_v$ . A Moebius strip is not orientable, so it has an area but no flux.
Area needs tangent planes, not only points: Schwarz’s lantern shows inscribed polyhedra can fail to converge to the area.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

The picture: patches, normals and flux

The idea, in words

The cross product

Cross product

Length and direction of the cross product

Parametric surfaces

Parametric surface

The fundamental vector product

Fundamental vector product

Normal and area element of a parametrized surface

Normal of a graph

Surface area

Surface area

Integrating a scalar over a surface

Surface integral

Orientation and flux

Oriented surface

Orientable surface

Flux

Flux as a double integral

The integral does not depend on the parametrization

Surface integrals do not depend on the parametrization

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere