Let $R$ be the area laying beneath the curve $f(x) = 8-x^2$ and above the line $y=7$. Find the volume of the solid of revolution which is created when $R$ is revolved around the $x$-axis.
I graphed the functions and found out I have to integrate from $-1$ to $1$. I want to use the disc method, but I don't know how to only get the discs with radii higher than $y=7$ and lower than $f(x)$. I thought it would be logical if the radii of the discs would be $8-x^2 -7 = 1-x^2$, but this approach gives me the wrong answer when i plug it into the formula.
Best Answer
Notice that the region you're supposed to rotate around the $x$-axis does not touch the $x$-axis. The solid that you get after rotation will therefore have a hole through it, centered around the $x$-axis.
What is called the "disc" method is usually used for solids without a hole, because the discs cross the axis of rotation and prevent the existence of a through-hole along that axis. When there is such a hole, as in your problem, the "washer" method is more frequently used. A "washer" is a disk with a circular hole in the middle, so it is described by two separate radius values: the outer radius of the "washer" and the radius of the hole. Instead of the $\pi r^2$ formula you use for a disk, you have $\pi (r_1^2 - r_2^2),$ where $r_1$ is the outer radius and $r_2$ is the radius of the hole.
In your case the outer radius is $8-x^2$ and the hole has radius $7.$
Alternatively, you can use the disc method to get the volume of the object obtained by rotating the area between the lines $x=-1$ and $x=1,$ the $x$-axis, and the curve $y=8-x^2$ around the $x$-axis, and then subtract the volume of the cylinder that is included in that object and not included in the object you were supposed to measure. This is equivalent to the washer method, because $\int\pi (r_1^2 - r_2^2)\,dx = \int\pi r_1^2 \,dx - \int\pi r_2^2\,dx.$