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Use the Divergence Theorem to compute the surface integral?

Use the Divergence Theorem to compute the surface integral where Q is bounded by z=x^2+y^2 and z=4, F=<x^3, y^3-z, xy^2>. (Answer: 32pi)

The divergence of the vector field is 3x^2+3y^2, which I've found.

How should I set up the triple integral and find the points of the integral and solve for it?

And where did you get 2*pi*r from?

Comments

z = x^2+y^2 = r^2

r = sqrt(z)

dxdy = rd?dr

Integrating d? from 0 to 2pi gives dxdy = 2pi*r dr

? = theta. I don't know why it was changed to "?"

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1 Answer

div F = ∂(x^3)/∂x + ∂(y^3-z)/∂y + ∂(xy^2)/∂z = 3x^2 + 3y^2

By the Divergence Theorem ∫∫F⋅N dS = ∫∫∫div F dV,

the surface integral

= ∫∫∫3(x^2+y^2) dx dy dz

= ∫[0,4]∫[0, sqrt(z)] 3(r^2) 2pi*r dr dz

= ∫[0,4] (3/2) pi z^2 dz

= (1/2) pi z^3 from 0 to 4

= 32 pi <==Answer