Using the shell method, find the volume of the solid by rotating the region bounded by the given curves

Question

$$x=y^2+1$$ $$x=2$$ about y=-2

How would I set this up? This is what I have so far:

$$V = \int_0^2 2 \pi (y+2)(y^2+1) dy$$

I am almost certain this is wrong. Especially with the limits of integration.

Answer 1

To get the integral right, we need a reasonably good picture. We have a rightward opening parabola with axis of symmetry the $x$-axis, and vertex at $(1,0)$. Note that the line $x=2$ meets our parabola at $y=\pm 1$.

Take a thin strip at height $y$, of width "$dy$," and rotate about the line $y=-2$. We get a cylindrical shell of radius $2+y$. The "height" of the cylinder is $2-x$, that is, $1-y^2$. This is because the strip extends rightward from $(x,y)$ to $(2,y)$. So the shell has approximate volume $2\pi(2+y)(1-y^2)\,dy$.

"Add up" (integrate) from $y=-1$ to $y=1$. The volume is $$\int_{y=-1}^1 2\pi(2+y)(1-y^2)\,dy.$$