Question

Use Stokes' Theorem to evaluate the surface integral where F= and where S is the portion of z=1-x^2-y^2 above the xy-plane with n upward. (Answer: -pi)

Answer 1

Haven't started Calc III yet =/

You are done Calc II right?

@Idealist

Answer 2

I hate these questions because it's ambiguous what they mean by \(\mathbf F\)

Answer 3

\[
\iint_S (\nabla \times \mathbf F) \cdot d\mathbf S = \int_{\partial S}\mathbf F\cdot d\mathbf s
\]

Answer 4

See the ambiguity is whether our vector is post-curl or pre-curl.
In this case, I'm assuming \(\mathbf F\) is going to be pre-curl.

Answer 5

@Idealist Can you draw the surface?

Answer 6

This looks like a circle. x^2+y^2=1, where r=1. So x=cos t, y=sin t, z=0 and dx=-sin t, dy=cos t, dz=0. And take the integral of zx^2 dx+z(e^xy^2)-x dy+x*ln(y^2) dz and I got 0 as the answer, which doesn't match the answer in the book. So can someone please correct me?

Answer 7

Don't they want you to take the curl instead?

Answer 8

The curl is <2y/y^2-1, 0, ze^y^2-1> but what do I do with that?

Answer 9

You can integrate over any surface with the same boundary.

Answer 10

that is another kicker of stokes theorem.

Answer 11

But how do I set up the integral with the limits?

Answer 12

Is this parametrisation correct though? I do believe that all points corresponding in the relation to the radius are important, so shouldn't it be:
x=rcost
y=rsint
z=1-r^2

while r ranges from 0 to 1 and t ranges from 0 to 2 pi

Answer 13

@Spacelimbus He parametrized the boundary, not the actual surface.

Answer 14

I need to look up on that then, seems complicated though, not suggesting that it's wrong then.

Answer 15

But should I find the dS?

Answer 16

@Idealist What did you get for \(\mathbf F\circ \mathbf r(t)\)?

Answer 17

I wanna see a step at a time.

Answer 18

What's r(t)?

Answer 19

Your parametrization of the boundary in this case.

Answer 20

I really don't know how to find that.

Answer 21

You have to find it to do a line integral... composing the function with the parametrization... You gotta know how to do that.

Answer 22

\[
\mathbf F(x(t),y(t),z(t))
\]

Answer 23

Oh, and you find dx, dy, dz with the given, right?

Answer 24

Yes. I want to see your work.

Answer 25

Wait a minute.

Answer 26

dx=2xz, dy=ze^2yx, dz=0.

Answer 27

Nope, \(dx = x'(t)dt\)

Answer 28

Just compose this function with the parametrizaiton first... can I see that?\[
\mathbf F=<zx^2, z(e^xy^2)-x, x*ln(y^2)>
\]

Answer 29

No, it's z(e^(xy^2)-x),

Answer 30

\[
\mathbf F=<zx^2, z(e^{xy^2}-x), x\ln(y^2)>
\]

Answer 31

dx=2xz dt
dy=ze^(2yx) dt
dz=0 dt

Answer 32

Wrong.

Answer 33

Are you finding the gradient? Why?

Answer 34

Sorry, I'm really confused by what you are trying to do.

Answer 35

As I said before, I hate these problems, because they somewhat specific to the book.

Answer 36

@Idealist Can you just ignore the whole \(dx,dy,dz\) part and just compose the vector field with the parametrization first? Please?

Answer 37

Yes, I'll try. I think I need to look up online to see the procedure because my book might be wrong, sorry to make you confused. And thank you for the help.

Answer 38

To do the composition, you have to replace the \(x,y,z\) with their respective functions of \(t\). There should be no \(x,y,z\).