Evaluating $\int_{-4} ^4\int _0 ^{\sqrt{16-x^2}} \int _0 ^{16-x^2-y^2} \sqrt{x^2 + y^2}\,dz\,dy\,dx$

Question

Question: Evaluate the given triple integral with cylindrical coordinates:

$$\int_{-4} ^4\int _0 ^{\sqrt{16-x^2}} \int _0 ^{16-x^2-y^2} \sqrt{x^2 + y^2}\,dz\,dy\,dx$$

My solution (attempt): Upon converting the triple integral into cylindrical coordinates, I got:

$$\int_{0} ^{\pi/2}\int _0 ^{4} \int _0 ^{16-r^2\cos^2\theta-r^2\sin^2\theta} r^2\,dz\,dr\,d\theta$$

After solving for the solution, I got $1024\pi/15$ as the answer. Apparently, this is incorrect. Could someone please show me where integral conversion is wrong?

Thanks in advance!

Answer 1

First integral should be from $0$ to $\pi$ because $-4 \le x \le 4$ and $0 \le y \le \sqrt{16 - x^2}$ is a semicircle. A full circle is $0$ to $2\pi$ and half of that is $0$ to $\pi$.