Question

stokes thoerem

Answer 1

http://gyazo.com/14c2a30d670ed3998895c619cb306446

Answer 2

So Stokes lets you change it into a flux integral.

Answer 3

Which means you would first find the curl of F.

Answer 4

\[
\int_C\mathbf F\cdot d\mathbf r = \iint_S\nabla \times \mathbf F\cdot d\mathbf S
\]

Answer 5

In this case \(S\) will be the surface \(z=y+1\).

Answer 6

Cylinder cutting through a plane.

Answer 7

So first find the curl of \(F\). Then parametrize the \(z=y+1\) surface.

Answer 8

Is curl F <0,-4x-4,-2> ?

Answer 9

I mean -4x + 4 for j

Answer 10

haven't calculated it myself... but it should be easy

Answer 11

Been a while since I have done any of this and I am trying to refresh my memory. :)

Answer 12

So, I hope you don't mind if I watch to see if I am remembering correctly.

Answer 13

I'm getting \(\nabla \times \mathbf F = (-4x+4)\mathbf j +(- 2)\mathbf k\).

Answer 14

Yes, that is what I got as well

Answer 15

<0, -4x+4, -2>

Answer 16

The normal vector for our parameterization for \(z=y+1\) should be \(z_x\mathbf i+z_y\mathbf j-\mathbf k\)

Answer 17

So I get \(
d\mathbf S = \mathbf j -\mathbf k
\)

Answer 18

So <0,-4x+4,-2> dot <0,1,-1>

Answer 19

or -4x + 6 ?

Answer 20

The limits of integration are \(x^2+y^2=1\), which translates to \(-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}\) and \(-1\leq x \leq 1\).

Answer 21

Could you switch to polar? r between 0 and 1 and theta from 0 to 2 pi?

Answer 22

and change x to rcos theta?

Answer 23

You can, but remember that \(x=r\cos\theta\). Does that make things much easier?

Answer 24

I don't know if it would make it any easier, actually, since you do end up with a y^2 when you do the integral, so the square root isn't so obnoxious

Answer 25

Because frankly, \(2x\sqrt{1-x^2}\) isn't that bad of an integral either.

Answer 26

Yes, you're right.

Answer 27

Hmmm, oh but there is the 6 term.. which will make a trig sub. I guess polar coords could help here.

Answer 28

I think I am just lazy and run for polar coordinates every time I see the equation of a circle or ellipse :)

Answer 29

Yeah, you are right. The 6 would give 6y when integrated, and that would get us stuck with the square root again

Answer 30

So, yes, maybe polar is the easier way to go

Answer 31

If you didn't want to use polar, you'd sub \(x=\sin\theta\).

Answer 32

Thank you! I appreciate your assistance. :)

Answer 33

Yes, you could go with trig substitution too.

Answer 34

I thought I'd try to see if it was doable using F dot dr and parameterizing the curve of intersection. That is ugly. :)

Answer 35

I don't know if I am remembering correctly, but would the boundary curve be the curve of intersection?

Answer 36

But curl F is much nicer than F itself.

Answer 37

Thank you !!!

Answer 38

The boundary curve is the intersection between the plane and the cylinder.