Stokes Theorem paraboloid intersecting with cylinder

gtfitzpatrick

1. The problem statement, all variables and given/known data

Use stokes theorem to elaluate to integral [itex] \int\int_{s} curlF.dS[/itex] where [itex] F(x,y,z)= x^2 z^2 i + y^2 z^2 j + xyz k [/itex] and s is the part of the paraboliod [itex]z=x^2+ y^2 [/itex] that lies inside the cylinder [itex] x^2 +y^2 =4 [/itex] and is orientated upwards

2. Relevant equations



3. The attempt at a solution

so i use Stokes theorem [itex] \int\int_{s} curlF.dS = \oint_{c} F.dv[/itex]

so i want to get parametric equations for C and so i get x=2cost y=2sint and z=4 i came up with these as the boundary curve c is a circle of raadius 2 and my z value of 4 because that is where the parabaloid and cylinder intersect...am i right in my thinking here?

so then dx= -2sintdt ; dy=2costdt and dz=0

so then i get [itex] \oint_{c} F.dv = \oint_{c} x^2 z^2 dx - y^2z^2 dy + xyzdz[/itex]

which becomes [itex]\int^{2\pi}_{0} ((4cos^2 t )(16)(-2sint) - (4sin^2t)(16)(ccost)) dt[/itex]

[itex]\int^{2\pi}_{0} ((-128cos^2 t )(sint) - (128sin^2 t)(cost)) dt[/itex]

am i working on the right lines here?

gtfitzpatrick

whoops hold up a sec...not finished putting up my attempt, im editing the post!

gtfitzpatrick

the origional post is complete now. anyone got any ideas?

gtfitzpatrick

anyone?

sharks

I think the last part is wrong and it should be:
##\int^{2\pi}_{0} (-128cos^2 t sint + 128sin^2 tcost) .dt##

LCKurtz

Check that - sign.

Looks good except for that minus sign.

sharks

The integration can be tricky but there is a simple trick to it. Let us know how it goes.

However, i have a question myself, since i'm still learning. I suppose it is OK for me to ask here? What if the orientation of the surface S was downwards? Then, how would the calculations from post #1 change?

gtfitzpatrick

Hi LCkurtz and Sharks
Thanks a million for comments.

To integrate i used substitution u=cost du=-sintdt and similar for sin.

as in [itex]\int cos^2 t sint =\frac{1}{3} cos^3 t[/itex] i think this works

so i get [itex]\frac{128}{3} (cos^3 t + sin^3 t)^{2\pi}_{0}[/itex]

=-[itex]\frac{128}{3}[/itex] i hope :)

gtfitzpatrick

i'm afraid i havent a clue, sorry!

sharks

But isn't the whole integral this:
##\int^{2\pi}_{0} (-128cos^2 t sint + 128sin^2 tcost) .dt##
Then, it would become:
##\int^{2\pi}_{0} (-128cos^2 t sint).dt + \int^{2\pi}_{0} (128sin^2 tcost).dt##

LCKurtz

I was tempted in my original post to ask the OP if he had thought about the orientation or whether he just took a guess. You use the right hand rule. In this case the normal was pointed upwards, so the right hand rule would give you counterclockwise in the xy plane as viewed from above which gives ##t:\, 0\rightarrow 2\pi##. For downward orientation you go the other way around the circle which could be ##t:\, 2\pi\rightarrow 0##. The effect is to change the sign of the answer.

sharks

Thank you for this quick and accurate answer, LCKurtz.

gtfitzpatrick

[itex]\int^{2\pi}_{0} (-128cos^2 t sint)dt + \int^{2\pi}_{0} (128sin^2 t cost)dt[/itex]

=128[[itex]\frac{1}{3} cos^3 t ]^{2\pi}_{0} + 128[\frac{1}{3} sin^3 t ]^{2\pi}_{0}[/itex]

=[itex]\frac{128}{3}[-1] = \frac{-128}{3} [/itex]

gtfitzpatrick

ahhh thanks a million i had wondered about that alright. that cleared it up. thanks a million.

sharks

I don't think the evaluation part is correct.

gtfitzpatrick

whoops should be [itex]\frac{128}{3}[1-1] = 0 [/itex]

gtfitzpatrick

Thanks a million sharks! your brill!