Question

Use cylindrical coordinates to find the volume of the solid. Solid inside \(x^{2}+y^{2}+z^{2}=16\) and outside \(z=\sqrt{x^{2}+y^{2}}\)

Short answer

The volume of the solid is \(V= \frac{32}{3}\pi\)

Step-by-step solution

Transform into cylindrical coordinates

Transform the equations of the sphere and cone into cylindrical coordinates. For the sphere: \(r^2+z^2=16\). For the cone: \(z=r\).

Define the limits of integration

Determine the limits of integration. Here, \(r\) goes from: \(0 \leq r \leq 4\), \(\theta\) goes from: \(0 \leq \theta \leq 2\pi\), and \(z\) goes from: \(r \leq z \leq \sqrt{16 - r^2}\).

Set up the integral

Set up the integral for volume using cylindrical coordinates. The volume \(V\) can be expressed as: \(V = \int_0^{2\pi} \int_0^4 \int_r^{\sqrt{16-r^2}} r\,dz\,dr\,d\theta\).

Calculate the integral

Calculate the integral. Doing so, the volume \(V\) is found to be: \(V= \frac{32}{3}\pi\).