Question

Use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations. $$ z=\sqrt{x^{2}+y^{2}}, z=0, x^{2}+y^{2}=25 $$

Short answer

The volume of the solid is \(\frac{250\pi}{3} cubic units\).

Step-by-step solution

Convert to polar coordinates

Given that the equation \(x^{2}+y^{2}=r^{2}\) represents a conversion from Cartesian to polar coordinates, we substitute \(x=r \cos(\theta)\) and \(y = r \sin(\theta)\) in our original equations. This gives us the new equations \(z=\sqrt{r^{2}}=r\), \(z=0\) and \(r=5\), in polar coordinates.

Determine the limits of integration

The radial coordinate \(r\) spans from 0 to 5, given by the circle equation \(r=5\). The angular coordinate \(\theta\) spans the full range of angles 0 to \(2\pi\), because the solid forms a full revolution about the z-axis.

Setup the double integral in polar coordinates

We are to find the volume of the solid. For this, we integrate the function \(z = r\), with respect to the area element in polar coordinates \(\mathrm{d}A = r\, \mathrm{d}r\, \mathrm{d}\theta\), over the domain described by the limits found in the previous step. Thus, the setup for the double integral becomes: \(\int_{0}^{2\pi}\int_{0}^{5} r\cdot r\, \mathrm{d}r\, \mathrm{d}\theta = \int_{0}^{2\pi}\int_{0}^{5} r^{2}\, \mathrm{d}r\, \mathrm{d}\theta\)

Evaluate the inner integral

The inner integral is over \(r\), and its integral is given by \(\int r^{2}\, \mathrm{d}r = \frac{1}{3}r^{3}\), which we evaluate from limits of 0 to 5. The result is \(\frac{1}{3}5^{3}-\frac{1}{3}(0) = \frac{125}{3}\). So the double integral now becomes \(\int_{0}^{2\pi} \frac{125}{3}\, \mathrm{d}\theta\).

Evaluate the outer integral

The remaining integral is \(\int_{0}^{2\pi} \frac{125}{3}\, \mathrm{d}\theta = \frac{125}{3}\theta\), evaluated at the limits from 0 to \(2\pi\). We get \(\frac{125}{3}(2\pi)-\frac{125}{3}(0) = \frac{250\pi}{3}\).