Finding the volume bounded by two equations rotated around $y = 2$

Question

I need to find the volume of a solid generated by a rotating the area bounded by the equations about $y = 2$:

• $y = 1$
• $y = x^2$

I've graphed the functions, but I'm not sure how to setup an integral to find the volume.

Answer 1

Think of this as a washer problem, with the volume resembling a ring. The inner radius is bounded by $y = 1$ while the outer is bounded by $y = x^2$. Where they intersect, at $x^2 = 2$ or $x = \sqrt 2$ is your upper bound of integration.

Notice that the inner radius is $2 - 1$ because that's the difference between $y = 2$ and $y = 1$. The outer radius is $2-x^2$ accordingly. Thus:

$$\pi\int_0^{\sqrt2} (2 -1)^2 - \left(2-x^2\right)^2 dx$$

Depending on what your lower bound is (it could be $x = -\sqrt 2$), you can multiply accordingly via symmetry (by $2$).