I want to find the volume of the region given by the intersection of the cone $x^2 + y^2 = z^2$ and the cylinder $y^2 + z^2 = 4$. I've tried the following (assuming only positive coordinates so I multiply the result by $8$): integrate with respect to $z$. When $z \leq \sqrt{2}$, the cone is not cut thus we can just consider the volume of a cone. For $\sqrt{2} < z \leq 2$, we have to integrate the area which is given by $\int_0^{\sqrt{4 - z^2}} \sqrt{z^2 - y^2} dy$, which corresponds to $1/4$ of the circle of radius $z$ cut by the cylinder boundaries on $y$. Thus the whole volume becomes $$\frac{4}{3} \pi \sqrt{2} + 8 \int_{\sqrt{2}}^2 \int_0^{\sqrt{4 - z^2}} \sqrt{z^2 - y^2}dy dz \; .$$ Image for clarification
The first term is twice the volume of each half-cone up to height $\pm \sqrt{2}$, and the second the integral on the area of the cone region bounded by the cylinder at some $z$. I've tried a change of coordinates $z \to 2 \sin{\theta}$ but it didn't help much. I would like to ask for help to evaluate the second integral. Any help is appreciated.
The Cartesian approach is not tractable in closed form, yielding an elliptic integral. Note that the correct volume is specified by $$V = \frac{4\sqrt{2}}{3} \pi + 8 \int_{z=\sqrt{2}}^2 \int_{y=0}^{\color{red}{\sqrt{4-z^2}}} \sqrt{z^2 - y^2} \, dy \, dz.$$ The issue is with the upper limit of integration highlighted in red; upon the substitution $y = z \sin \theta$, it becomes $$\theta = \arcsin \sqrt{\frac{2}{z^2} - 1}.$$
A better approach is to use cylindrical coordinates with respect to the $x$-axis; i.e., $$z = r \cos \theta, \quad y = r \sin \theta, \quad x = x,$$ hence the integrand in the circular sector becomes $$\sqrt{z^2 - y^2} = r \sqrt{\cos 2\theta}$$ and the desired volume becomes $$V = 8 \int_{r = 0}^2 \int_{\theta = 0}^{\pi/4} r^2 \sqrt{\cos 2\theta} \, d\theta \, dr$$ after accounting for the Jacobian of the transformation: $$dV = r \, d\theta \, dr \, dx.$$ Then since the above is separable, $$V = \frac{32}{3} \int_{\theta=0}^{\pi/2} \sqrt{\cos \theta} \, d\theta.$$ At this point we know that $\sqrt{\cos \theta}$ has no elementary closed-form antiderivative, thus confirming the claim I stated at the beginning. However, the definite integral does have a closed form in terms of the gamma function: $$\int_{\theta = 0}^{\pi/2} \sqrt{\cos \theta} \, d\theta = \sqrt{\frac{2}{\pi}} \Gamma^2 (\tfrac{3}{4}).$$ This of course is nothing more than a rephrasing of the beta function integral $$\int_{z=0}^1 z^{a-1} (1-z)^{b-1} \, dz = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = 2 \int_{\theta=0}^{\pi/2} \sin^{2a-1} \theta \cos^{2b-1} \theta \, d\theta$$ for the choice $(a,b) = (1/2, 3/4)$. This gives a numeric value $$V \approx 12.780162503846316879\ldots.$$