Question

I'm trying to use Stokes Theorem to find the Flux of the Curl; never done a problem quite like this, posting below momentarily.

Answer 1

http://i.imgur.com/oKolx8F.png

Answer 2

Up until now, I've always been dealing with Stokes' Theorem in the form of calculating line integrals from looking at a 2D object and finding its' area with grad cross F dot n as the integrand; I've never done this before now, specific prompt listed below:

http://i.imgur.com/9aFreml.png

Answer 3

@ganeshie8 , could you help me on this possibly?

Answer 4

It's funny, do they want you to use stokes theorem to change it to a flux of curl problem, or do they want you to use it to make it a line integral problem?

Answer 5

\[
\oint_C \mathbf F\cdot d\mathbf r = \iint_S\nabla\times\mathbf F\;\cdot d\mathbf S
\]Where \(C\) is the boundary of the surface \(S\).

Answer 6

I'm assuming they want you to use the right side, since \[
\nabla \times \mathbf F = -2x\mathbf j
\]

Answer 7

Do you know how to calculate curl?

Answer 8

Whoop! Sorry, I went and started working on some other stuff, I'm back. One second, taking a look.

Answer 9

I'm not sure. These problems are in a separate subsection from the ones where you took a line integral problem and then put it as a double integral.

Answer 10

Darn, the second post I linked to the wrong photo, one sec.

Answer 11

http://i.imgur.com/pvHgsos.png

Answer 12

That is the original prompt.

Answer 13

I hate these "use stokes theorem" problems...

Because they they imply going from one way to another when they really want you to just do the original problem.

Answer 14

Anyway, as I said before, Stoke's theorem says:
\[
\color{red}{\oint_C \mathbf F\cdot d\mathbf r }= \color{blue}{\iint_S\nabla\times\mathbf F\;\cdot d\mathbf S}
\]Where \(C\) is the boundary of the surface \(S\).

Answer 15

The original problem is the \(\color{blue}{\text{blue}}\) part, so I assume we want go convert to the \(\color{red}{\text{red}}\) part.

Answer 16

Btw, how did your color your LaTeX?

Answer 17

Also, how did you do upright text?

Answer 18

And yeah, I think we want to convert to the red part and evaluate the circulation/line integral.

Answer 19

Yuh

Answer 20

So it would seem that \(C\) is given by: \[
(x,y,0),\quad 0\leq x\leq a,\quad 0\leq y\leq \sqrt{a^2-x^2}
\]

Answer 21

Hold on, let me think about that for a moment...

Answer 22

\[
\mathbf r(t) = (a\cos(t),a\sin(t),0)
\]

Answer 23

I'm changing it to polar coords to keep things simple. \(t\in [0,2\pi)\)

Answer 24

\[
\mathbf F\circ \mathbf r = -a\sin(t)\mathbf i+a\cos(t)\mathbf j+a^2\cos^2(t)\mathbf k
\]

Answer 25

\[
d\mathbf r = (-a\sin(t), a\cos(t),0)dt
\]

Answer 26

\[
\mathbf F\cdot d\mathbf r = [a^2\sin^2(t)+a^2\cos^2(t)+0]dt = a^2\;dt
\]

Answer 27

I think this leaves us with:\[
\color{red}{\oint_C \mathbf F\cdot d\mathbf r }= \int_0^{2\pi}a^2\;dt

\]

Answer 28

I'm taking a look at everything, one moment. Thank you already for the help.

Answer 29

I'm just confused about the phrasing of the whole thing; Stokes' Theorem is used for when you have a single line integral or a sum of line integrals you are calculating, but in this situation...I dunno, the phrasing is so bad, I can't definitively find in the prompt the point where they choose to use the top of the cylinder or the bottom to be what you calculate as the basis of the line integral, or both, or anything else.

Answer 30

Which part have I said that you didn't know where it came from?

Answer 31

One at a time.

Answer 32

The first bit where you said, (x, y, 0); we implicitly decided that the bottom of the cylinder, the curve connecting that bottom region to the cylinder body was our line integral basis. I'm not saying you're wrong, I think you're right, but the prompt confused me and I couldn't understand what you saw in the prompt that made it clear that we were using the curve along the bottom of the surface, as opposed to the top of the surface

Answer 33

That is the boundary of the surface in this case.

Answer 34

You don't have a boundary at the top, because the top is still part of the surface.

Answer 35

Also, I know you changed them, but I don't understand why the original cartesian bounds' lower limits aren't just the negative of their upper bounds

Answer 36

That was a mistake.

Answer 37

The parametrization shouldn't have contained \(y\) to begin with, because it should be a single variable parametrization.

Answer 38

If I had kept it in Cartesian coordinates, it would have been problematic because I would have had to split it up into two semi circles.

Answer 39

"Let S be the cylinder x^2 + y^2 = a^2, 0 \leq z \leq h, together with its top, x^2 + y^2 \leq a^2, z = h."

It's a piecewise-defined surface, I don't understand how the top isn't a boundary or a separate part, like, if I had a right circular cylinder with a smooth, piecewise-defined body with the top, bottom, and middle defined, I could have two potential stokes' theorem problems, right? One at the bottom and one at the top along the cap ends of the cylinder.

Answer 40

oh yeah, no, I got why you changed, that would've been super messy, was just making sure I wasn't missing something

Answer 41

Oh, wait a minute...

Answer 42

Does this cylinder just......not have a bottom? Like an opened can? There isn't a bottom, is there.

Answer 43

If you were to look at a cross section of the cylinder, you would get: