If the circle $(x - b)^2 + y^2 = a^2 (0 <a< b)$ is revolved about the y-axis, it generates a doughnut-shaped solid called a torus. Find the volume of this torus by the washer method.
Let's use the washer method, the upper half of the circle is
$$x = \sqrt{a^2-y^2} + b $$
the lower half: $$x = b - \sqrt{a^2-y^2}$$
then:
$$dV = \pi[(\sqrt{a^2-y^2} + b)^2-(b - \sqrt{a^2-y^2})^2]dy = 4b\sqrt{a^2-y^2}$$
what are the limits of integration?
$$V = \int_0^{a+b}4b\sqrt{a^2-y^2}dy$$?