Edit: I realized where my mistake was...thanks for the help!
In class we were discussing a problem (page 1185 in Smith and Minton's Calculus Third Edition):
Let Q be the solid bounded by the paraboloid $z=4-x^2-y^2$ and the $xy$-plane. Find the flux of the vector field $\mathbf F(x,y,z) = \big <x^3,y^3,z^3 \big>$ over the surface $\partial Q$.
We know from the Divergence Theorem that the flux of $\mathbf F$ over $\partial Q$ is given by
$\displaystyle \iint_{\partial Q}\mathbf F \cdot \mathbf n \ dS = \iiint_Q \nabla \cdot \mathbf F(x,y,z) \ dV$
The book then evaluates the right side to be $96\pi$. In class we tried to confirm by evaluating the left side of the equation, but we went wrong somewhere, so my question is could someone help me with the left side or tell me where I went wrong. My work so far:
The surface is composed of the paraboloid part and the disc part so I broke up the surface integral into two integrals each over one part.
The disc: $\displaystyle \iint_{x^2+y^2 \leq4}\mathbf F \cdot \mathbf n \ dS$
but the normal points straight down so it will be $\big <0,0,$ something$\big >$, which when you dot it with $\mathbf F$ will be $\big <0,0, z^3$ something$\big >$, but since z=0 the whole integral is zero, so all we have to worry about is the paraboloid part.
Paraboloid part:
We can parameterize it as follows:
$x = r\cos\theta \; y=r\sin\theta \; z=4-r^2$ where $0 \leq r \leq 2$ and $0\leq \theta \leq 2\pi$.
Crossing partials gives a normal of: $\big <2r^2\cos\theta, 2r^2\sin\theta,r \big >$
Dotting with $\mathbf F = \big <r^3 \cos^3\theta, r^3\sin^3\theta, (4-r^2)^3 \big>$ and putting into the integral gives
$\displaystyle \int_0^{2\pi}\int_0^2 [2r^4\cos^4\theta + 2r^4\sin^4\theta + (4-r^2)^3 r]r \ dr \ d\theta$ I'm pretty sure the terms with cosine and sine will cancel out because either they will have sine which will be zero when 0 and $2\pi$ are plugged in or they will have cosine and the subtracting will cancel them out (if the two terms at each endpoint are the same), so all we have to worry about is the right most term meaning:
$\displaystyle 2\pi \int_0^2 64r-48r^3+12r^5-r^7 dr$
$\displaystyle 2\pi[32r^2-12r^4+2r^6-\frac{r^8}{8}]_0^2 = 64\pi$
Where did I go wrong?