Computing the integral $\iiint_D\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}~\mathrm{d}x~\mathrm{d}y~\mathrm{d}z$

Question

Computing the integral

$$\iiint_D\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}~\mathrm{d}x~\mathrm{d}y~\mathrm{d}z$$

where $D$ is the portion of the ball $x^2 + y^2 + z^2 \leq 16$ that satisfies $z \geq 2$.

(a) Evaluate this integral using cylindrical coordinates.

(b) Evaluate this integral using spherical coordinates.

The solution I found was $\frac{8\pi}{3}$ as answer for both, cylindrical and spherical. Cylindrical coordinates

$\iiint\limits_D \frac{z}{(x^2+y^2+z^2)^{3/2}} dV = \int\limits_{0}^{4}\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{2\pi}\frac{\rho^2 \cos{\phi} \sin{\phi}}{\rho^3} \sin{\phi} d\rho d\phi d\theta = \frac{8\pi}{3}$

[Espherical coordinates][2] [2]: https://i.stack.imgur.com/KvUEC.png \begin{align*} \int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{4} \frac{\rho \cos\phi \sin\phi}{\rho^3} \rho^2 \sin\phi ,d\rho,d\phi,d\theta &= \int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{4} \cos\phi \sin^2\phi ,d\rho,d\phi,d\theta \ &= \int_{0}^{2\pi} \int_{0}^{\pi/3} \left[\frac{\rho\cos\phi \sin^2\phi}{-2}\right]0^4 ,d\phi,d\theta \ &= \int{0}^{2\pi} \int_{0}^{\pi/3} -8\cos\phi \sin^2\phi ,d\phi,d\theta \ &= \int_{0}^{2\pi} \left[\frac{2\cos^3\phi}{3}\right]0^{\pi/3} ,d\theta \ &= \int{0}^{2\pi} \frac{2}{3} \left(1 - \frac{1}{8}\right) ,d\theta \ &= \frac{8\pi}{3}. \end{align*}

Then, it also gives $\frac{8\pi}{3}$.

Answer 1

You have computed the integral$$\int_0^{2\pi}\int_0^{\pi/3}\int_0^4\frac{\rho^2\cos(\varphi)\sin(\varphi)}{\rho^3}\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta,$$which is not the integral that you should be computing, since it corresponds to the integral of the given function over $D$ together with the cone whose vertex is the origin and whose base is the circle at the bottom of $D$. So, you are integrating of a larger region than $D$.

You can compute your integral using cylindrical coordinates:\begin{align}\int_0^{2\pi}\int_2^4\int_0^{\sqrt{16-z^2}}\frac{z\rho}{(\rho^2+z^2)^{3/2}}\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta&=2\pi\int_2^41-\frac z4\,\mathrm dz\\&=\pi.\end{align}

If you want to, you can use spherical coordinates:\begin{align}\int_0^{2\pi}\int_0^{\pi/3}\int_{2\sec(\varphi)}^4\cos(\varphi)\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta&=2\pi\int_0^{\pi/3}4\cos(\varphi)\sin(\varphi)-2\sin(\varphi)\,\mathrm d\varphi\\&=\pi.\end{align}