Finding the volume of a solid of revolution about the x-axis 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.
 A: 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
$\int\pi (r_1^2 - r_2^2)\,dx = \int\pi r_1^2 \,dx - \int\pi r_2^2\,dx.$
A: Notice that you want to integrate the area of each disk, not the radii. So the radius of the outer disk is $8-x^2$ and thus the area inside is $(8-x^2)^2$. From there see if you can do it.
A: You can think it as subtracting the solid of revolution of $g(x)=7 $ (from -1 to 1),from the solid of revolution of$f(x) $from (-1 to 1).You get the wrong answer because you calculate the solid of revolution of $h(x)=1-x^2$ from -1 to 1 wich is a different solid.
