I have an area $R$, which lies beneath the curve $y = 9 - x^2$ and over the line $y = 5$. What is the volume when I rotate $R$ around the $x$-axis?
This is what I've done so far:
First I wanted to find out the integration limit, so I put $9 - x^2 = 5$, and found $x_1 = -2$ and $x_2 = 2$.
I then tried to find the inverse, and ended up with $x = \sqrt{y + 9}$.
This is probably wrong, but I think the radius is $5 + y$.
So I took the integral of $2 \pi(5+y)(2-\sqrt{y + 9})$ from $-2$ to $2$, but got the wrong answer. What did I do wrong? Thanks for the help.
You can express the volume of this solid as the difference of the volumes of two solids, since the area in question can be expressed as a difference of two areas: the volume of the solid formed by rotating the area under $y=9-x^2$ from $x=-2$ to $x=2$ about the $x$-axis, and the volume of the solid formed by rotating the area under $y=5$ from $x=-2$ to $2$ about the $x$-axis. The latter solid of rotation is simply a cylinder with radius $5$ and height $4$, so its volume is $100\pi$. The volume of the former solid is given by $$\int_{-2}^2 \pi(9-x^2)^2dx$$ and so the volume you seek is the difference, given by $$\int_{-2}^2 \pi(9-x^2)^2dx-100\pi$$