Volume of the region of sphere between two planes.

Question

I want to find the volume of the region of the sphere $x^2+y^2+z^2=1$, between the planes $z=1$ and $z=\frac{\sqrt{3}}{2}$

I have used triple integral for calculating this $$\int _0^{2\pi }\int _{0}^{\frac{\pi }{6}}\int _{0 }^1\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta $$

Are the limits of integral i have chosen correct??

Edit: As from the comment, my integral is not correct, so please clarify what region actually the above integral represents.

Answer 1

The bottom of your region is a horizontal plane which is not the $xy$-plane. This means you should at least consider cylindrical coordinates. For instance, $$ \int_0^{2\pi}\int_0^{1/2}\int_{\sqrt3/2}^{\sqrt{1-r^2}} r\,dz\,dr\,d\theta $$ or $$ \int_0^{2\pi}\int_{\sqrt3/2}^1\int_0^{\sqrt{1-z^2}}r\,dr\,dz\,d\theta $$ In case you really want spherical coordinates, if we put the radius as the innermost integral, note that it doesn't go from $0$, it goes from wherever $z=\sqrt3/2$, which is to say from $\frac{\sqrt3}{2\cos\phi}$, which is not easy to integrate. If we do $\phi$ first instead, we get $$ \int_0^{2\pi}\int_{\sqrt3/2}^1\int_0^{\arccos(2\rho/\sqrt3)}\rho^2\sin\phi\,d\phi\,d\rho\,d\theta $$ which might look scary at first, but everything turns out pretty nice in the end.

Answer 2

You do not even need calculus to solve this one. If you visualize, this is a sphere with center at (0,0,0) and radius 1. So the plane z = 1, is tangent to its surface at (0,0,1). The plane $z = \frac{\sqrt3}{2} \lt 1$ is parallel to XY plane and will cut the sphere into a spherical cap. As you have to find the volume of sphere between these two planes, you really have to find the volume of the spherical cap.

Volume of spherical cap = $\frac {\pi h^2(3R-h)}{3}$ where h is the height of the spherical cap and R is the radius of the sphere. $R = 1, h = (1-\frac{\sqrt3}{2})$. Solving it, you get the volume as $\frac {\pi}{3}(2-\frac{9}{8}\sqrt3)$.