<s>[INST] Evaluate a triple integral with spherical coordinates $$\iiint_{Q}z\ dV$$ Where $Q$ is the common region of the spheres $x^{2}+y^{2}+z^{2}\leq 1$ and $x^{2}+y^{2}+(z-1)^{2}\leq 1$
I have tried nothing.
 [/INST] We have to use spherical coordinates $(x=r\sin\theta\cos\varphi, \quad y=r\sin\theta\sin\varphi, z=r\cos\theta)$, so we will have $$x^2+y^2+z^2 \le 1 \Leftrightarrow r^2 \le 1 \Leftrightarrow r \le 1$$ and $$x^2+y^2+(z-1)^2 \le 1 \Leftrightarrow x^2+y^2+z^2 \le 2z \Leftrightarrow r^2 \le 2r\cos\theta \Leftrightarrow r \le 2\cos\theta.$$
Next, since both spheres have their centers on $z$-axis, one center is $P_1=(0,0,0)$ and another one is $P_2=(0,0,1)$ we know that it will hold: $$0 \le \varphi\le 2\pi,$$ and since both spheres have radius $1$ it is not hard to see that it will hold: $$0 \le \theta \le \frac{\pi}{2}.$$
Finally, your integral will be equal to $$\int_{0}^{2\pi}d\varphi \int_{0}^{\frac{\pi}{2}}\sin\theta \cos\theta d\theta \int_{2\cos\theta}^{1}r^3 dr=2\pi\int_{0}^{\frac{\pi}{2}}\sin\theta \cos\theta(\frac{1}{4} - 4\cos^4\theta)d\theta =...= \frac{-13\pi}{12}.$$
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