Hello, softballchick!
I think I understand the problem . . .
The region enclosed by $\displaystyle y = x^2,\;y = 3x$ is revolved about the line $\displaystyle x = 0.$
Find the volume using washers.
You didn't show any work, but I assume your setup was way off.
The graph looks like this:
Code:

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They wanted the volume found by "washers".
. . So we must work "sideways".
The two functions are: .$\displaystyle \begin{Bmatrix}x_{_R} &=& y^{\frac{1}{2}} \\ \\[3mm]x_{_L} &=& \frac{1}{3}y \end{Bmatrix}$
$\displaystyle \displaystyle\text{The formula is: }\;V \;=\;\pi\int^b_a\bigg[\left(x_{_R}}\right)^2  \left(x_{_L}\right)^2\bigg]\,dy $
$\displaystyle \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 $
$\displaystyle \displaystyle \text{So we must evaluate: }\:V \;=\;\pi\int^9_0\left(y  \tfrac{1}{9}y^2\right)dy$
Got it?
Edit: Too slow ... again!
.