∫∫(∇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.
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.
Similar to S1, I can think of this hemisphere as the circle
x²+y² = 9 on the x-y plane with z = 8 right?
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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