# Use triple to verify that a paraboloid divides a solid in two regions of the same volume, where am I wrong?

Let $S$ be the region over the $xy$ plane and inside the intersection of the cylinder $x^2+y^2=a^2$ and the plane $z=a^2$.

I want to verify that the paraboloid $z=x^2+y^2$ divides $S$ into two solids of equal volume.

I did the following:

Let the first region, $A_1$, be the one inside the paraboloid. One can use cylindrical coordinates to find out its volume, which is:

$$V(A_1)=\int_0^{2\pi} \int_0^{a} \int_{r^2}^{a^2} r\, dz \, dr \, d\theta=2\pi a^4(\frac{1}2-\frac{a}3)$$

Let the second region, $A_2$, be the one outside the paraboloid and inside the cylinder. Then, using the same coordinates its volume is:

$$V(A_2)=\int_0^{2\pi} \int_0^{a} \int_{0}^{r^2} r\, dz \, dr \, =\frac{\pi}2 r^4$$

Where I am going wrong? I just haven't found out what my mistake is. Thanks for your help.

Your integrals are set up correctly (except that the second one is missing $d\theta$ — I assume it's just an oversight). Both integrals evaluate to $\pi a^4$, so your second evaluation was pretty close, but the first one came out weird. It's hard to tell what might have gone wrong.