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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!

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  • $\begingroup$ HINT: Use spherical coordinates, but replace the second sphere with the sphere $x^2+y^2+(z-R)^2 = R^2$ centered on the $z$-axis. It has a very nice equation in spherical coordinates: Using US conventions on $\phi,\theta$, it is $\rho = 2R\cos\phi$. Now, you will need to split it into two integrals, one with $0\le\phi\le\pi/3$ and one with $\pi/3\le\phi\le\pi/2$. The $\rho$ limits will be different in the two integrals, of course. Draw a picture. $\endgroup$ – Ted Shifrin Aug 13 at 17:33
  • $\begingroup$ See the amended answer below for an illustration of triple integral for this problem. $\endgroup$ – Quanto Aug 13 at 20:14
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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,

$$V=2 \int_{R/2}^{R} \pi y^2dx=2\pi\int_{R/2}^{R} (R^2-x^2)dx=\frac{5}{12}\pi R^3$$

enter image description here


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.

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  • $\begingroup$ I don't get your order of integration, you say integrate first over z, but after integration 'dz' is still in the integration, did you mean integrate first over x. If so how did you find the lower boundary for the x component? $\endgroup$ – Eliot Janssens Aug 14 at 7:45
  • $\begingroup$ Right, over x. Lower bound is from solving x of the 2nd curve. $\endgroup$ – Quanto Aug 14 at 11:06

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