Question

Stokes' Theorem

I'm trying to evaluate the line integral around the curve obtained by intersecting the cylinder x^2+z^2=1 with the plane y=3. The vector field is F=<3xz, e^(xz), 2xy> and the curve is oriented counterclockwise when viewed from the right.

Answer 1

I parameterized the surface needed to use stokes' thm as r(u,v)=<u*cos(v), 3, u*sin(v)> where 0 <=u<=1 and 0<=v<=2*pi. The curl of F and the cross product of the partials of r(u,v) are quite long, but I got an answer of 6*pi which is far from the answer in the solution key.

Answer 2

F(r) . r' right?

Answer 3

For the line integral it is, but we are supposed to use Stokes' theorem

Answer 4

i can never keep the names straight

Answer 5

what is stokes thrm by chance?

Answer 6

Flux is F.n

Answer 7

The line integral in space is equal to the flux of the surface

Answer 8

\[\int\limits_{C} F.dr = \int\limits \int\limits_{S} curl(F).dS\]

Answer 9

what was your normal vector? <0,y,0? ?

Answer 10

oy, its way to late for me to be doing anything remotely sane with that :/

Answer 11

If the plane intersects the cylinder you know the cross section's shadow on the x-y plane is a circle right? So you need only to integrate around the circle? And since it's a surface integral r if FIXED at r=1 not to be integrated over. So integrate over z and phi (the polar angle).

Answer 12

I believe.

Answer 13

Especially with y=3, its not even slanted so it will DEFINITELY only give you a circle of constant radius.

Answer 14

The formula I have been following states that flux integral over a paramterized surface is \[\int\limits \int\limits_{D} F(r).(r_u \times r_v) dA\] This, in effect, eliminates having to compute the normal directly.

Answer 15

And since you're only integrating with scalars you don't need to worry about integrating over a "position" vector which would NOT give you a circle if the plain was slanted.

And you still compute the normal directly, you just don't normalize it.

Answer 16

And I guess it would be the x-z plane not x-y.

Answer 17

Integrating along the curve would likely be more simple, but we are supposed to use Stokes' theorem instead of directly computing the line integral

Answer 18

but the normal is quite obvious though, isn't it? It just points in the y-direction.

so <0,1,0> is fine

Answer 19

Yes, I meant we don't compute the nomalized normal, the wording gets tricky ;)

Answer 20

And I misread, since you want to calculate the line integral you only need to integrate over phi. So:

So:
\[\Phi(\phi)=(- \cos(\phi),3,\sin(\phi))\]
That is your parameterization for that circle.
So
\[0 \le \phi \le 2 \pi\]
And then you have:
\[\oint \vec{F} \cdot d \vec{l}=\int\limits_0^{2 \pi} \vec{F}(\vec{\Phi}(\phi)) \cdot \vec{\Phi}'(\phi)d \phi\]