Doubt with bounds and integrand of $\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{2}\rho^2\sin{\phi}d\rho d\phi d\theta$

Question

Question as follows.

Find the volume of the solid enclosed between the spheres $x^2+y^2+z^2=4$ and $x^2+y^2+z^2=4z \Leftrightarrow x^2+y^2+(z-2)^2=4$.

I constructed the following integral and after computation got the answer $\frac{16\pi}{3}$:

$$\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{2}\rho^2\sin{\phi}d\rho d\phi d\theta=\frac{16\pi}{3}.$$

The book says the answer should be $\frac{10\pi}{3}$ so I'd like to know what I did wrong. Are my bounds wrong? Should my integrand be different and not just the Jacobian?

Thank you.

Answer 1

Your integral's bounds are not nearly that simple. All you've integrated over is a ball that goes from the origin to a radius of 2. That doesn't describe the intersection of these two spheres. What you've done is like saying "I have a triangle whose corners are the origin, (0,1) and (1,0)" therefore my integral's bounds are $\int_0^1 \int_0^1 \, dx \, dy$. That's not right; you'd get twice the area of the triangle. You can't just take the max and min of each coordinate and integrate willy nilly.

What you can do, however, is use some symmetry. The circle of intersection between the two spheres defines a plane that cuts through the spheres. You can find the volume of the spherical sector defined by that plane restricting $\phi$ to some range $[0, \phi_0]$ (you will have to find the opening angle of the spherical sector to do this). You can then subtract off the volume of the cone that is cut off by the plane of intersection. What remains is the part of the sphere that is cut off by the plane of intersection. Since the geometry is symmetric, you can use this for the other sphere as well, and you're done.

Answer 2

Hint: I think it'd be easier to approach this in cylindrical coordinates.

Thus, try out the integral $$\int\limits_0^{2\pi}\int\limits_0^{\sqrt{3}}(2\sqrt{4-r^2}-2)\,r\,dr\,d\theta$$

Answer 3

A related problem. You should plot the two spheres to see what's happening. Note that, the upper bounds of $\rho$ are different and the solution is

$$ \int_{0}^{2\pi}\int_{0}^{\pi/3}\int_{0}^{2}\rho^2\sin{\phi}\,d\rho\, d\phi\, d\theta+ \int_{0}^{2\pi}\int_{\pi/3}^{\pi/2}\int_{0}^{4\cos(\phi)}\rho^2\sin{\phi}\,d\rho\, d\phi\, d\theta=\frac{10\pi}{3}.$$