The volume of the region between two spheres and the upper nappe of a cone Guys I am really having trouble constraining the region between these three surfaces. I am imagining a sort of "Dome", or a "muffin head" sort of shape. Is this correct ? Anyway, I need to be able to write the following volume integral in rectangular, cylindrical and spherical coordinates: Consider the region that is between $x^2 + y^2 + z^2 = 1$, $x^2 + y^2 + z^2 = 9$, and finally above the upper nappe of the cone $z^2 = 3(x^2 + y ^2)$ upon further consideration, does the smaller sphere even matter ? wouldn't it just represent a hole in the larger sphere, an area im not even worried about finding the volume of ? Anyway, thanks for looking!

First find where the cone intersects the inner sphere:

$$x^2+y^2=1-z^2$$

$$z^2=3-3z^2$$

$$z^2=\frac{3}{4}$$

This means that the radius of the boundary between the cone and inner sphere is $\frac{1}{2}$, and the radius of the boundary between the cone and the radius of the boundary for the outer sphere can be found to be $\frac{\sqrt{33}}{2}$, using the same process. Converting the sphere equations into cylindrical coordinates and using as bounds $0 \le \theta < 2\pi$ and $\frac{1}{2}\le r < \frac{\sqrt{33}}{2}$, the volume integral is as follows:

$$V=\int_0^{2\pi}\int_{\frac{1}{2}}^{\frac{\sqrt{33}}{2}}(\sqrt{9-r^2}-\sqrt{1-r^2})rdr d\theta$$

A similar procedure can be followed to find the integral in rectangular and spherical coordinates.