Help with specific Volume Integral related to divergence theorem I would like to solve the volume integral
$\int_V div(F)dV$ where $F = (x^3 + 3y + z^2, y^3, x^2 + y^2 + 3z^2) $ and $0\le z\le1, 1-z = x^2 + y^2$
I don't know what limits to use. I'm thinking $(1-x^2)^\frac 12, 0$ and same for $y$, and then $0, 1$ for $z$
Thanks!
 A: By the Divergence Theorem we know
$\displaystyle \iiint_Q \nabla \cdot \mathbf F\ dV = \iint_{\partial Q} \mathbf F \cdot \mathbf n \ dS $
so I will compute this integral via the right side.
Parameterizing the surface $z = 1-(x^2+y^2), 0 \leq z \leq 1$ as
$\mathbf r(r,\theta) = \big <r\cos\theta, r\sin\theta, 1-r^2 \big>$
we can now find the normal by crossing the partials.
$\mathbf r_r = \big <\cos\theta, \sin\theta, -2r \big>$
$\mathbf r_\theta = \big <-r\sin\theta, r\cos\theta, 0 \big>$
$\mathbf n = \mathbf r_r \times \mathbf r_\theta = \big <2r^2 \cos\theta, 2r^2\sin\theta, r \big>$
[Note: we don't have to worry about normalizing this since the magnitude will cancel out when we convert $dS$ into $dA$ via $dS = \| \mathbf r_r \times \mathbf r_\theta \| dA$.]
Now dotting n with F and plugging back into the integral gives:
$\displaystyle \int_0^{2\pi} \int_0^1 [2r^5+6r^3\cos\theta\sin\theta+r^3+(1-r^2)^2(2r^2\cos\theta+3r)] \ r \ dr \ d\theta$
which after some laborious but simple integration gives $\frac{10\pi}{7}$ (you can confirm with Wolfram Alpha but if you do it by hand you'll get the satisfaction of many cancellations :) )
