Question

Use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations. Inside the hemisphere \(z=\sqrt{16-x^{2}-y^{2}}\) and outside the cylinder \(x^{2}+y^{2}=1\)

Short answer

The volume of the solid is \(15\pi\) cubic units.

Step-by-step solution

Define the Limits of Integration

The given equations for the hemisphere and the cylinder are symmetrical about both the x and y axes, so we have to integrate the function only over the first quadrant and then multiply it by 4. The radius r should vary from 1 (the smallest radius defined by the cylinder) to 4 (the edge of the hemisphere), and the angle \(\theta\) should vary from 0 to \(\pi/2\).

Set Up the Double Integral

In polar coordinates, the double integral is given by the formula \(\int \int r dz dr d\theta\). We replace z in the formula by \(\sqrt{16-r^{2}}\) which is obtained from the equation of the hemisphere in polar form. The limits for r are from 1 to 4, and for theta are from 0 to \(\pi/2\). So the double integral is \(\int_{0}^{\pi/2} \int_{1}^{4} r\sqrt{16-r^{2}}drd\theta\).

Evaluate the Inner Integral

The inner integral is evaluated first. Here, we have to perform a substitution: let \(u=16-r^{2}\), so \(du=-2r dr\). After this substitution, the integral will be easier to solve.

Evaluate the Outer Integral

After obtaining an expression from the inner integral, we proceed to evaluate the outer integral with the limit from 0 to \(\pi/2\).

Multiply by 4

Since the original integral was calculated over the first quadrant only, multiply the final result by 4 to get the volume of the entire solid.