Evaluate:

$\displaystyle \iiint_V (x+y+z)dV$ where V is the region bounded by the paraboloid $\displaystyle z = 4-x^2-y^2$ and the $\displaystyle xy$-plane.

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My sol'n (but I am unsure, because it's ugly!):

[Ah, just realised I missed an 'r' when I was integrating as I wrote this. The solution is now simpler, but the computation is very long]

When z=0, the region R in the xy-plane is the circle with radius=2. So $\displaystyle R=\{(x,y)\mid x^2+y^2\le 4\}$ so in polar coords,

$\displaystyle 0\le r \le 2$ and

$\displaystyle 0 \le \theta \le 2\pi$ and

$\displaystyle 0\le z \le 4-r^2$.

So we have: (I integrated along Z first, don't know if that's right)

$\displaystyle \int_0^{2\pi} \int_0^2 \int_0^{4-r^2} (r\cos\theta + r\sin\theta + 4-r^2) r dz dr d\theta$

Skipping boring computation (the second time I did it in maxima

), we get:

$\displaystyle \frac{64\pi}{3}$.

Is that right?

EDIT: Hm... looking at the integrand now, I'm thinking I should have left the z in there, and not substituted it for 4-r^2...