Evaluating $\int_{-4} ^4\int _0 ^{\sqrt{16-x^2}} \int _0 ^{16-x^2-y^2} \sqrt{x^2 + y^2}\,dz\,dy\,dx$
By John Campbell •
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!
$\endgroup$1 Answer
$\begingroup$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 $\pi$.
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