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.

  • $\begingroup$ Btw your using of MathJax is looking more understandable to me than your image 🤣. However both are understandable. Good to see you using MathJax and showing your work. Keep uo the good work and have a upvote from me $\endgroup$ – Jitendra Singh Jun 5 at 7:08
  • $\begingroup$ Use $ \int \sqrt{a^2-x^2} = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin(\frac{x}{a})$, when integrating with respect to $y$. $\endgroup$ – Vedant Chourey Jun 5 at 7:21

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 $$V \approx 12.780162503846316879\ldots.$$


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