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Stokes's Theorem problem

  • Thread starter izelkay
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  • #1
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Homework Statement


http://puu.sh/d5FpW.png [Broken]

Homework Equations


∫∫(∇XF)⋅dS = ∫F(r(t))⋅r'(t)dt

The Attempt at a Solution


So I figured I'd have two line integrals (using Stokes's theorem.

Paramatizing S1:

I can think of the cylinder as the closed curve circle x²+y² = 9 on the x-y plane right?

x = 3cost, y = 3sint, z = 0; 0 ≤ t ≤ 2pi

or r(t) = (3cost)i + (3sint)j

Then F = (27(cost)^3)i +(2187(sint)^7)j

The dot product of F and r'(t) gives me

-81sint(cost)3 + 6561(sint)^7cost

Then I integrate this from t = 0 to t = 2pi and get 0.

For S2:

Similar to S1, I can think of this hemisphere as the circle

x²+y² = 9 on the x-y plane with z = 8 right?

Then paramatizing:

r(t) = (3cost)i + (3sint)j + 8k ; 0≤t≤2pi

I go through the same process as S1, but I don't get 0 I get -576 pi, which is obviously wrong.

Would someone mind walking me through how to do this problem correctly?Am I applying Stokes's theorem incorrectly.
 
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Answers and Replies

  • #2
LCKurtz
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The only boundary of the silo/hemisphere is the circle in the xy plane, and you got zero for that.
 
  • #3
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Hi LCKurtz, thanks for your reply.

I was thinking I may only need to do the integral for S1 since, as you said, both surfaces share a circle in the xy plane as the boundary, but thought that made the problem too trivial.

So instead of splitting the surface into two parts S1 and S2, I can think of it as one surface like the middle image in this picture:
image001.png

so what I calculated for S1 is for the entire surface as well?
 
  • #4
LCKurtz
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Yes. One way to think of it is if that surface was flexible rubber or a soap bubble, it could contract down to just the disk inside that circle. Any way you deform the surface still has that boundary.
 
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