Volume when rotated about the line $y=-1$

Question

Find the volume when the region enclosed by $y=x^2$, $y=4$ is revolved around the line $y=-1$

My teacher has given the following answer:

$$\text{D.} \quad 4\pi \int_0^4(y+1)\sqrt{y} \, dy$$

I assume she has done this through the method of shells, due to the the $4\pi$ outside the integral, but I don't know how to come up with this. Can anyone help? All I know is that $\sqrt y$ is when $y$ is a function of $x$, and the $y+1$ is the radius I guess? But where does the $4\pi$ come from?

Answer 1

It is clear that the radius of the shell is $r = y + 1$ (the height of the blue line minus the height of the red line). The height of the shell is given by

$$ h = \sqrt y - (-\sqrt y) = 2 \sqrt y.$$

Using cylindrical shells, the desired quantity is

$$ \int_0^4 2\pi r h \ dy = \int_0^42\pi(y+1)(2\sqrt y) \ dy = 4 \pi \int_0^4(y+1) \sqrt y \ dy \approx 228. $$

See the plot below: