# Question about Solid of Revolution (Disc Method)

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• Mar 2nd 2011, 07:49 AM
Lemonie
Question about Solid of Revolution (Disc Method)
So, I know the formula for the disc method is: $v=pi\int\limits_{b}^{a}[R(x)]^2dx$ ...what I'm so confused about is how to fix up $[R(x)]$ when the axis of revolution isn't the x-axis or the y-axis. For example:

http://i52.tinypic.com/nqhp5.jpg

What is the equation to be used when using the disc method to get:
a.) shaded region when revolved to y=8
b.) non-shaded region when revolved to y=8

Thank you in advance!
• Mar 2nd 2011, 08:09 AM
HallsofIvy
A disk will have radius extending from the graph to the line of rotation. If the line of rotation is y= 8, then the radius is R(x)= |y- 8|= |f(x)- 8|. For B, the non-shaded area, since it goes up to the line y= 8, the volume would be
$\pi\int_{x=0)^{x_1} (f(x)- 8)^2 dx$
where the upper limit, $x_1$ is the value of x where the graph of y= f(x) crosses the line y= 8. That is, $f(x_1)= 8$.

For (a) where there is unshaded area between the shaded area and the line of rotation, you cannot use the "disk method" directly. What you can do is use the "washer" method. A washer has an outer radius of $r_2$ an inner radius of $r_2$ and so its area is the area between those two radii, $\pi r_2^2- \pi r_1^2= \pi(r_2^2- r_1^2)$. That is really the same as doing two separate "disk" calculations, to find an "outer volume" and "inner volume" and subtracting:
$\pi\int_a^b R_2(y)^2dy- \pi\int_a^b R_1(y)^2dy$

For a problem like the one you show, where one edge is just y= 0, that outer volume is just the volume of a cylinder- $\pi R^2 h$ which here would be $64\pi h$ and, again, h is the value of x where the graph crosses the line of rotation, f(x)= 8.