Recently I've been proposed the next problem at my calculus course.
Find the volume of the following corps: $$\left\{(x,y,z) \in \mathbb{R}^3 : \; x^2+y^2+z^2 \le 2(y+z), \; x^2+y^2 \le z^2\right\}$$
On first place, I attempted to plot the functions via Mathematica. This is a screenshot of the plot
I've thought to divide the figure in 4 shards like this and this, so I can find the volume of each one separately. (sorry for only having the images, I don't know how to explain it)
For the first part, I tried to find the intersection of the sphere ($x^2+y^2+z^2≤2(y+z)$) and the cone ($x^2+y^2 \le z^2$) and try to find a parabolic cylinder that goes trough that intersection and is normal to the XY plane, so I can have $y$ in function of $x$ so I'm able to do the integral, but I'm struggling here, so I hope you can help me.
The volume projection onto the $yz$-plane is the region enclosed by $z=y$ and $y^2+z^2=2(y+z)$, obtained by setting $x=0$. It is partitioned by the intersection projection of the two shapes, i.e. $z^2=(y+z)$. Then, convert to cylindrical coordinates with $(y,z)\to(r, \theta)$. The bounds are $\theta \in(\frac\pi4,\frac{3\pi}4)$, $r_s(\theta)= 2(\sin\theta +\cos\theta ) $ and the partition is $r_p(\theta)= \frac{\sin\theta +\cos\theta }{\sin^2\theta}$. The sphere and the core are respectively $$x_s(r,\theta)=\sqrt{2r(\sin\theta+\cos\theta)-r^2} ,\>\>\>\>\>x_c(r,\theta)= r\sqrt{\sin^2\theta-\cos^2\theta} $$ and the volume integral is \begin{align} \int_{\frac\pi4}^{\frac{3\pi}4}& \int_0^{ r_p(\theta)}2x_c(r,\theta) rdrd\theta +\int_{\frac\pi4}^{\frac{3\pi}4} \int_{r_p(\theta) }^{r_s(\theta )} 2x_s(r,\theta) rdrd\theta= \frac{7\pi}{12}+\frac{11\pi}{12}=\frac{3\pi}2 \end{align} where the first integral is for the cone and the second for the sphere.