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I have to calculate the volume of intersection of a sphere and a cylinder. The cylinder's radius is $r$ and the center point is $(r,0,0)$. The sphere's center point is $(0,0,0)$ and the radius $2r$.

I calculated the estimated volume with Monte Carlo methods but now I have to calculate the real volume with a formula. Is there any kind of explicit equation for this volume?

I will appreciate any kind of help!

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    Please show that estimated volume calculation and any and all other relevant info, please! – 2017-01-10
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    ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3; – 2017-01-10
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    I used this to calculate the estimated, now I have to calculate the theoretical one. – 2017-01-10

2 Answers 2

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Yes, draw a figure, and you will realize that the volume is given by $$ 2\iint_D \sqrt{(2r)^2-x^2-y^2}\,dx\,dy $$ where $$ D=\{(x,y)\in\mathbb R^2~|~(x-r)^2+y^2\leq r^2\}. $$ I leave it to you to calculate the integral.

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    Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :( – 2017-01-10
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Let the sphere be $x^2+y^2+z^2 = 4a^2$ And the cylinder be $(x - a)^2+y^2 = a^2$ Using polar coordinates we have $$x = r cos\theta cos\phi$$ $$y = r sin\theta cos\phi$$ $$z = r sin\phi$$ Then the equation of the sphere is r = 2a And the cylinder $r cos\phi = 2a cos\theta$ Solving the 2 equations we have the curve of intersection $cos\phi = cos\theta$ or $\phi = \theta$ This means that for a fixed $\theta$ we have $\phi$ varies from 0 to $\theta$ and r varies from 0 to 2a and $\theta$ from 0 to $\pi$ / 2 by considering only the first quadrant. By transforming to polar coordinates we have the volume integrant to be $$\frac{D(x, y, z)}{D(\theta, \phi, r)} = r^2cos\phi$$ Hence the total volume is $$4\int_0^\frac{\pi}{2}\int_0^{2a}\int_0^\theta r^2cos{\phi}d{\phi}drd{\theta}$$ $$= 4\int_0^\frac{\pi}{2}\int_0^{2a}r^2sin{\theta}drd{\theta}$$ $$= 4\int_0^\frac{\pi}{2}\frac{8a^3}{3}sin{\theta}d{\theta}$$ $$= \frac{32a^3}{3}$$