Answer
$2\pi^2 a^2b$
Work Step by Step
Step 1. Draw a diagram as shown in the figure. We have the circle as $x^2+y^2\leq a^2$ and a line $x=b$, where $b\gt a$.
Step 2. With the circular region revolving around $x=b$, using cylinder approximation and the symmetry, we have
$V=2\int_{-a}^a 2\pi (b-x)(\sqrt {a^2-x^2})dx=4\pi b\int_{-a}^a (\sqrt {a^2-x^2})dx-4\pi \int_{-a}^a x(\sqrt {a^2-x^2})dx=4\pi b(\frac{\pi a^2}{2})+2\pi \int_{-a}^a (a^2-x^2)^{1/2}d(a^2-x^2)=2\pi^2 a^2b+\frac{4\pi}{3} (a^2-x^2)^{3/2}|_{-a}^a=2\pi^2 a^2b+(0-0)=2\pi^2 a^2b$
Step 3. Instead of using cylinders and integrating over $dx$, we can also use washers and integrate over $dy$ as
$V=2\int_{0}^a \pi (r_2^2-r_1^2)dy=2\pi \int_{0}^a ((b+\sqrt {a^2-y^2})^2-(b-\sqrt {a^2-y^2})^2)dy=2\pi \int_{0}^a (2b(2\sqrt {a^2-y^2}))dy=8\pi b(\frac{\pi a^2}{4})=2\pi^2 a^2b$
