Verify Stokes' theorem

Verify Stokes' Theorem for the vector field F= and the cylinder x^2+y^2=9 for 1≤ z ≤2 oriented out.


Both integrals should equal 54π.

(i) Surface integral.

Since curl F = <2xz, 2yz, -2z^2>, and a parameterization of the cylinder is
r(u, v) = <3 cos u, 3 sin u, v> for u in [0, 2π], v in [1, 2] with r_u x r_v = <3 cos u, 3 sin v, 0>:

∫∫s curl F · dS
= ∫∫s <6v cos u, 6v sin u, -2v^2> · <3 cos u, 3 sin v, 0> dA
= ∫(v = 1 to 2) ∫(u = 0 to 2π) 18v du dv
= ∫(v = 1 to 2) 36πv dv
= 18πv^2 {for v = 1 to 2}
= 54π.
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(ii) Line integral(s):
The top curve C₁: x^2 + y^2 = 9, z = 2 is oriented clockwise as viewed from above, while
the bottom curve C₂: x^2 + y^2 = 9, z = 1 is oriented counterclockwise viewed from above.

So, we compute using polar parameterization x = 3 cos t, y = 3 sin t, z = (z-value).
∫c₁ F · dr + ∫c₂ F · dr
= -∫(t = 0 to 2π) <(3 sin t) * 2^2, -(3 cos t) * 2^2, 2^3) · <-3 sin t, 3 cos t, 0> dt
+ ∫(t = 0 to 2π) <(3 sin t) * 1^2, -(3 cos t) * 1^2, 1^3) · <-3 sin t, 3 cos t, 0> dt

(The minus sign in the first integral is for orientation, as the polar parameterization works counterclockwise.)

Evaluating this yields
-∫(t = 0 to 2π) -36 dt + ∫(t = 0 to 2π) -9 dt
= 54π.

I hope this helps!