# Finding volume using disc or washer

• March 9th 2011, 04:48 AM
softballchick
Finding volume using disc or washer
The volume of the solid obtained by rotating the region enclosed by about the line http://math.webwork.rochester.edu:80...13319ac2f1.png can be computed using the method of disks or washers via an integral
http://math.webwork.rochester.edu:80...cb8f222c01.png =

I got 86http://math.webwork.rochester.edu:80...1af403ed41.png but that's wrong... Also I cant find the boundaries. I know a=0 and when I graphed the functions, they cross at x=2 but that's not right....
• March 9th 2011, 05:41 AM
skeeter
$y = 3x$ ... $x = \dfrac{y}{3}$

$y = x^2$ ... $x = \sqrt{y}$

note that the two graphs intersect at y = 0 and y = 9

using the method of washers about the y-axis (that is the line x = 0) ...

$\displaystyle V = \pi \int_0^9 \left(\sqrt{y}\right)^2 - \left(\dfrac{y}{3}\right)^2 \, dy$
• March 9th 2011, 05:57 AM
Soroban
Hello, softballchick!

I think I understand the problem . . .

Quote:

The region enclosed by $y = x^2,\;y = 3x$ is revolved about the line $x = 0.$
Find the volume using washers.

You didn't show any work, but I assume your set-up was way off.

The graph looks like this:

Code:

       |     9+ - - - - - *       |        *:|       |      *::*|       |    *:::  |       |  *::::*  |       | *:::*    |     --*-----------+------       |          3

They wanted the volume found by "washers".
. . So we must work "sideways".

The two functions are: . $\begin{Bmatrix}x_{_R} &=& y^{\frac{1}{2}} \\ \\[-3mm]x_{_L} &=& \frac{1}{3}y \end{Bmatrix}$

$\displaystyle\text{The formula is: }\;V \;=\;\pi\int^b_a\bigg[\left(x_{_R}}\right)^2 - \left(x_{_L}\right)^2\bigg]\,dy$

$\displaystyle \text{We have: }\;V \;=\;\pi\int^9_0\bigg[\left(y^{\frac{1}{2}}\right)^2 - \left(\tfrac{1}{3}y\right)^2\bigg]\,dy$

$\displaystyle \text{So we must evaluate: }\:V \;=\;\pi\int^9_0\left(y - \tfrac{1}{9}y^2\right)dy$

Got it?

Edit: Too slow ... again!
.