Help calculating integral over a region $G$ with spherical coordinates?

Question

So i have to calculate this triple integral:

$$\iiint_GzdV$$ Where G is defined as: $$x^2+y^2-z^2 \geq 6R^2, x^2+y^2+z^2\leq12R^2, z\geq0$$

So with drawing it it gives this:

It seems i should use spherical coordinates since sphere is involved. But i don't know if i can use the default ones or should manipulate them since i have hyperbole of some sort(but since its $a=b=c$, is it a double sided cone then? i know that z>0 means im only looking at upper half, but still)

So using this:

$$x=r\cos\phi\cos\theta; y=r\sin\phi\cos\theta ; z=r\sin\theta$$

So i think $$\phi \in [0,2\pi]$$ And for $$r: r^2\leq12R^2$$ But am not sure about theta, how can i get it? From the $z\geq0$ i get $r\sin\theta>=0$ so $\sin\theta \geq 0$ So from that is $0\leq\theta\leq \frac{\pi}{2}$

Am I doing it correctly, or should i fix anything, thank you in advance.

Answer 1

The condition $$ x^2+y^2-z^2\geq 6R^2 $$ gives $$ r^2(\sin^2\theta-\cos^2\theta)\geq 6R^2 $$ So you want $(r,\theta)\in(0,\infty)\times(0,\frac\pi2)$ such that $$ r^2\leq 12R^2 \text{ and } r^2(1-\cos 2\theta)\geq 6R^2 $$ So $1-\cos 2\theta\geq\frac12$, i.e, $\cos 2\theta\leq \frac12$ and hence $\theta\in(0,\frac\pi6)$. Putting them together, $$ \\\int_G z\,\mathrm{d}V= \int_0^{2\pi}\int_0^{\pi/6}\int_{R\sqrt{6/(1-\cos 2\theta)}}^{R\sqrt{12}} r\cos\theta\,r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi $$

Answer 2

Wanting to calculate the following triple integral: $$I := \iiint\limits_{\Omega} z\,\text{d}x\,\text{d}y\,\text{d}z$$ with: $$\Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : x^2 + y^2 \ge 6R^2 + z^2, \; x^2 +y^2 + z^2 \le 12R^2, \; z \ge 0 \; ; \; R > 0 \right\},$$ opting for a transformation from rectangular to polar coordinates, we have: $$ \begin{cases} \rho^2\,\sin^2\varphi \ge 6R^2 + \rho^2\,\cos^2\varphi \\ \rho^2 \le 12R^2 \\ \rho\,\cos\varphi \ge 0 \\ \rho \ge 0 \\ 0 \le \varphi \le \pi \\ 0 \le \theta < 2\pi \end{cases} \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} \sqrt{\frac{6}{-\cos(2\varphi)}}\,R \le \rho \le 2\sqrt{3}\,R \\ \frac{\pi}{3} \le \varphi \le \frac{\pi}{2} \\ 0 \le \theta < 2\pi \end{cases} $$ from which: $$I = \int_0^{2\pi} \text{d}\theta \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sin(2\varphi)}{2}\,\text{d}\varphi \int_{\sqrt{\frac{6}{-\cos(2\varphi)}}\,R}^{2\sqrt{3}\,R} \rho^3\,\text{d}\rho = \frac{9}{2}\,\pi\,R^4\,, $$ while opting for a transformation from rectangular to cylindrical coordinates, we have: $$ \begin{cases} \rho^2 \ge t^2 + 6R^2 \\ \rho^2 + t^2 \le 12R^2 \\ t \ge 0 \\ \rho \ge 0 \\ 0 \le \theta < 2\pi \end{cases} \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} \sqrt{6R^2 + t^2} \le \rho \le \sqrt{12R^2 - t^2} \\ 0 \le \theta < 2\pi \\ 0 \le t \le \sqrt{3}\,R \end{cases} $$ from which: $$I = \int_0^{\sqrt{3}\,R} t\,\text{d}t \int_0^{2\pi} \text{d}\theta \int_{\sqrt{6R^2 + t^2}}^{\sqrt{12R^2 - t^2}} \rho\,\text{d}\rho = \frac{9}{2}\,\pi\,R^4\,.$$ Since $\Omega$ has a cylindrical and non polar symmetry, it is clear that the second way is less calculative.