Stokes’ Theorem Math 2400-005, Calculus III, Fall 2018

Stokes’ Theorem: Let S be an oriented piecewise-smooth surface that is

bounded by a simple, closed, piecewise-smooth boundary curve C with positive

orientation. Let ~F be a vector field whose components have continuous partial

derivatives on an open region in R3that contains S. Then

Z

C

~F · d~r =ZZ

S

curl ~F · d~S.

In this course, we will primarily see three applications of Stokes’ Theorem: using an

easy line integral to evaluate a di�cult surface integral, using an easy surface integral

to calculate a di�cult line integral, and using a simple surface integral to calculate a

di�cult surface integral.

There will be an example and then a similar exercise for each type. For the exercises,

I will give the correct answer in class so that you can check your work.

Example 1. Evaluate

ZZ

S

curl ~F · d~S where ~F = hy, x� 2x3z, xy3i and S is the upper

half sphere x2+ y2 + z2 = 1 oriented upwards.

Solution. Let’s take a moment to look at this surface and decide on the best course

of action.

The boundary of this region is the unit circle x2+ y2 = 1 on the xy-plane, i.e. when

z = 0. If we use Stokes’ Theorem and evaluate the line integral instead, then we have

z = 0 and the third component should vanish in the integrand. Also, we can avoid calcu-

lating the curl. This will simplify the problem considerably, so we’ll try the line integral.

We have that the boundary is x2+ y2 = 1 with z = 0, so now we can parametrize

this curve as r(✓) = h cos(✓), sin(✓), 0i.with 0 ✓ 2⇡. Recall that we are going to use

the formula Z

C

~F(r(✓)) · r0(✓) dt.

Thus we record that r0(✓) = h�sin(✓), cos(✓), 0i and that ~F(r(✓)) = h sin(✓), cos(✓), cos(✓) sin3(✓)i.

Now we’re ready to plug everything into our formula.

Z 2⇡

0

h sin(✓), cos(✓), cos(✓) sin3(✓)i · h� sin(✓), cos(✓), 0i d✓

=

Z 2⇡

0

cos2(✓)� sin

2(✓) d✓

=

Z 2⇡

0

cos(2✓) d✓

= 0.

Exercise 1. Evaluate

ZZ

S

curl ~F · d~S where ~F = h � y, x, zi and S is the part of the

sphere x2+ y2 + z2 = 25 above the plane z = 4.

Using the same trick, we'll evaluate this integral by doing a line integral around the

boundary curve .

Parameter iz ingC : X 't y 't 42=25 ⇒ xzty 2=9

Z = 4Fct ) = ( 3 cos HI ,

35in HI, 4) O Et I ZIT

FCF HD = f -

y , x , z ) =L-3 sine ) ,3 cos Lt )

, 4)F Kt ) = ( -3 sin HI

, 3 cos Lt ) , O )E ( Fut ) . F ' Lt ) = 9 sin At t 905kt ) to = 9

Since C is the boundary of S and is positively oriented , then by Stokes '

theoremS§ curl F. d 's = { to DEFECTED . fittldt -

- Si"9dt=l8④

Example 2. Evaulate

Z

C

~F · d~r where ~F(x, y, z) = hx � z, x + y, y + zi and C is the

intersection of the cylinder x2+ y2 = 1 and the plane y = z oriented counterclockwise.

Solution. Again, we want to take a moment and look at the curve from a couple of

perspectives. First, we’ll examine the cylinder stopping at the plane y = z:

Now, let’s look at the curve C by itself in space:

We could try to evaluate the line integral, but let’s not. Instead, we could use Stokes’

Theorem to evaluate a surface integral over the following surface:

We could parametrize this using the fact that y = z and otherwise we are going in a

circle. Since the projection onto the xy-plane is the disk x2+ y2 1, then our bounds

will be the same as the disk. Let’s do this explicitly now.

To parametrize our region, we note that we are forming a circle of radius 1 in the x and

y coordinates, so we have x = r cos(✓) and y = r sin(✓). We need to have r so that we

form a surface and not a curve. Now we recall that this was formed by intersecting with

the plane y = z, so the ellipse lies on the plane y = z and we get z = r sin(✓). Thus ourparametrization for our surface S is given by ~r(r, ✓) = hr cos(✓), r sin(✓), r sin(✓)i with0 r 1 and 0 ✓ 2⇡.

We’re at the point now where we should remind ourselves of the formula. To calculate

a surface integral, we use the formula:

ZZ

D

curl ~F(~r(r, ✓)) · (~rr ⇥~r✓) dA.

Note that you may not know whether the order of the cross product is correct. I checked

beforehand, so I used the ordering that gives me a positive orientation. However, if you

notice mid-problem that you did the wrong order, then make a quick comment on your

paper and multiply by �1.

To evaluate the integral, I should first find the cross product and the curl. We have

that

~rr = h cos(✓), sin(✓), sin(✓)i

~r✓ = h � r sin(✓), r cos(✓), r cos(✓)i

So we calculate the cross product

~rr ⇥~r✓ =

������

i j kcos(✓) sin(✓) sin(✓)

�r sin(✓) r cos(✓) r cos(✓)

������= h0,�r, ri

And we confirm that this vector is pointing in the positive direction which matches the

positive orientation of our curve. Now let’s calculate the curl of ~F:

curl ~F =

������

i j k@@x

@@y

@@z

x� z x+ y y + z

������= h1,�1, 1i.

Now we have enough information to finally evaluate the integral:

ZZ

D

h1,�1, 1i · h0,�r, ri dA

=

Z 2⇡

0

Z 1

0

2r dr d✓

=

Z 2⇡

0

1 d✓

= 2⇡.

Exercise 2. Evaulate

Z

C

~F · d~r where ~F(x, y, z) = hx+ y2, y + z2, z + x2i and C is the

triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) oriented counterclockwise.

This line integral would require 3 parts , but if we do a surface integral we can

get the job done with one integral .

What do we need ? I j K

• curl (E) = Ox E -

- det / Zx Zg Zz )= (O - Hi - ( 2x - o ) j t ( o -

zy) I

ixty ytz ztx = f Zz ,- 2x , -2g )

• A parameterization of the surface S : S is part of the plane Xtytz =L,

so we can use Flay ) = ( x , y , I - x -

y ) with O Exel, of ye I - x .

• A normal vector : In this case we know that ds-ftfxtcfyT-ftft.it -

- Bso we can use a unit normal vector

. From what we know about planes ,

D= 4 ,I

,I > is a normal vector and it = 4¥12 is a unit normal vector .

Carl (E) Circuit ) = f- 24 - x -

y ) ,-2C x )

, -2cg ) ) = ( - 2t2xt2y ,- 2x , -2g )

By Stokes ' theorem ,

£ E. die =

Sgs curl E) ads =

"

So - ztzxtzy ,- 2x

, -2g) . 41¥11 . Fsdydx

= So'S:X - ztzxtzy - 2x -2g dydz = Sis ! -2 dydx = So'

-2+2 xdx

= f 2x t xD ! = -2T I = ①

Example 3. Evaluate

ZZ

S

curl ~F ·d~S where ~F(x, y, z) = hxyz, xy, x2yzi where S is the

top and four sides of the cube with vertices (±1,±1,±1) oriented outwards.

Solution. Let’s start by taking a look at the surface. We’ll do a standard angle and

then we’ll look at the bottom of the cube, which is missing a side.

At this point we should feel mild to moderate nausea at the prospect of doing a surface

integral for each side. This unpleasant feeling is a good clue that we should use Stokes’

Theorem.

Unlike in Example 1, we aren’t going to evaluate a line integral. The reason being

doing a line integral over each side would only marginally be less of a pain.

Remember that as a consequence of Stokes’ Theorem, we can do a surface integral

over a simpler surface and get the answer, as long as it shares the same boundary. In

this case, the boundary is the bottom edges of the cube (the ones without a side).

This shares a boundary with the much simpler surface

This region is simply parametrized by ~r(x, y) = hx, y,�1i where �1 x 1 and

�1 y 1. Remember that we’ll be using the formula

ZZ

R

curl ~F(~r(x, y)) · (~rx ⇥~ry) dA.

And note that this time, the cross product is obviously h0, 0, 1i since this gives us the

normal vector. Now we need to calculate the curl:

curl ~F =

������

i j k@@x

@@y

@@z

xyz xy x2yz

������= hx2z, xy � 2xyz, y � xzi.

Putting this all together, we can now evaluate the integral

ZZ

R

h � x2, xy + 2xy, y + xi · h0, 0, 1i dA

=

Z 1

�1

Z 1

�1

x+ y dx dy

= 0.

Exercise 3. Evaluate

ZZ

S

curl ~F · d~S where ~F = hexy cos(z), x2z, xyi and S is the

hemisphere x =

p1� y2 � z2 oriented in the direction of the positive x-axis.

We can close up this hemisphere with the disc D : y't EE I in the yz

- plane ,so S and

D have the same boundary ( but with opposite orientation ).

Therefore by Stokes ' theorem

Sss curl tools = -

S§ cark Et od 5.

What do we need ?I j K

• curl CE ) = ox F = det ) Zx Ey Zz I = ( x . xD i - Cy - tends in GD) Jt ( 2x z - x eat costa ) I

eat costs) X' z XY

• A unit normal vector : Since D is in theyz

- plane ,

At-40,0 ) is a unit normal vector

oriented outward . And since D is flat , ds = IDA = dxdy .

Thus curl CES on DA -

- H- x ) dxdy = Odxdy since x -- O on D

.

Therefore S§ curl CEI eds -

- Sf O dxdy = O.