Math Help - Finding volume using disc or washer

1. Finding volume using disc or washer

The volume of the solid obtained by rotating the region enclosed by
about the line can be computed using the method of disks or washers via an integral
=

I got 86 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....

2. $y = 3x$ ... $x = \dfrac{y}{3}$

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

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) ...

$\displaystyle V = \pi \int_0^9 \left(\sqrt{y}\right)^2 - \left(\dfrac{y}{3}\right)^2 \, dy$

3. Hello, softballchick!

I think I understand the problem . . .

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!
.