Consider the vector field F=...(Calc 3) Picture of question attached

Here's the question:

http://imgur.com/mZUeG

correct answer/explanation will get you 10 points...thanks!

curl F = <-3x, -2y, 5z>.

Next, using the Cartesian (default) parameterization,
n = <-z_x, -z_y, 1> = <2x, 2y, 1>.

So, ∫∫s curl F · dS
= ∫∫ <-3x, -2y, 5(4 - x^2 - y^2)> · <2x, 2y, 1> dA
= ∫∫ (20 - 11x^2 - 9y^2) dA.

Convert to polar coordinates (noting that the region of integration is bounded by x^2 + y^2 = 3):
∫∫ (20 - 11x^2 - 9y^2) dA
= ∫(r = 0 to √3) ∫(θ = 0 to 2π) [20 - 11r^2 cos^2(θ) - 9r^2 sin^2(θ)] * r dθ dr
= ∫(r = 0 to √3) ∫(θ = 0 to 2π) [20 - 2r^2 cos^2(θ) - 9r^2] * r dθ dr
= ∫(r = 0 to √3) ∫(θ = 0 to 2π) [20 - r^2 (1 + cos(2θ)) - 9r^2] * r dθ dr
= ∫(r = 0 to √3) ∫(θ = 0 to 2π) [20r - 10r^3 - r^3 cos(2θ)] dθ dr
= ∫(r = 0 to √3) [2π(20r - 10r^3) - 0] dr
= π(20r^2 - 5r^4) {for r = 0 to √3}
= 15π.

I hope this helps!