Hi I have been looking for this triple integral almost a week now, I can not find a good integral at the end, I need to solve this with triple integrals, I know there are other ways to calculate it. Can someone solve this problem? I tried to solve it with cylindrical co-ordinates and spherical.
Let's have two spheres with radius R, one sphere his centre point is on the outer shell of the other, calculate the volume of the intersection of those two spheres, so for instance I took these two equations:
$$S1: x^2 + y^2 + z^2 = R^2$$
$$S2: (x-R)^2 + y^2 +z^2 = R^2$$
Thanks a lot!
The common volume is formed from a xy-area revolving around x-axis. The area is twice of that under the following curve due the same radius,
$$y^2+x^2=R^2$$
So, the volume can be simply obtained from the disk integration,
$$V=2 \int_{R/2}^{R} \pi y^2dx=2\pi\int_{R/2}^{R} (R^2-x^2)dx=\frac{5}{12}\pi R^3$$
Given the high symmetry of the problem, the triple integral, as illustrated below, may be an overkill. First, integrate over $x$,
$$V=\int_V dzdydx = \int_S dzdy\int_{R-\sqrt{R^2-y^2-z^2}}^{\sqrt{R^2-y^2-z^2}}dx=\int_S dzdy \left( 2\sqrt{R^2-(y^2+z^2)} - R \right)$$
For the resulting double-integral, it is convenient to use the polar coordinates due to circular boundary. Rewrite the integral as,
$$V=\int_0^{2\pi} \int_0^{\frac{\sqrt{3}}{2}R} \left( 2\sqrt{R^2-r^2 }- R \right) rdrd\theta$$ Since there is no $\theta$-dependency in the integrand, the integral simplifies to,
$$V=2\pi \int_0^{\frac{\sqrt{3}}{2}R} \left( 2\sqrt{R^2-r^2 }- R \right) rdr=\frac{5}{12}\pi R^3$$
which can be carried out by hand.