Find the volume $V$ of the solid below the paraboloid $z=8-x^{2}-y^{2}$ and above the following region. \[ R=\left\{(r, \theta): 1 \leq r \leq 2,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\right\} \] Set up the double integral, in polar coordinates, that is used to find the volume. (Type exact answers.) Find the volume. \[ \mathrm{V}= \] (Type an exact answer, using $\pi$ as needed.)
Solution
Step 1 ：We are given a solid below the paraboloid \(z=8-x^{2}-y^{2}\) and above the region \(R=\left\{(r, \theta): 1 \leq r \leq 2,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\right\}\). We are asked to find the volume of this solid.
Step 2 ：We can find the volume of the solid by setting up a double integral in polar coordinates. The limits of integration for r will be from 1 to 2, and for theta will be from -pi/2 to pi/2.
Step 3 ：The function to be integrated will be the function for the paraboloid, converted to polar coordinates. In polar coordinates, x = rcos(theta) and y = rsin(theta), so the function becomes z = 8 - r^2.
Step 4 ：The differential area element in polar coordinates is rdrdtheta, so the double integral becomes: \[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{1}^{2} (8 - r^2) r dr d\theta\]
Step 5 ：Evaluating this integral gives us the volume of the solid. The final answer is \(\boxed{\frac{33\pi}{4}}\)
