I am stuck on the following problem.
First I converted $y = \sqrt{3x}$ to $x = \frac{1}{3y^2}$. Then I tried to solve for volume using the following formula: $$ V = \pi \int_0^3 \left(\frac{1}{3} y^2\right)^2\,dy $$ But ultimately I got an answer of $3\pi$ which the solutions says is incorrect.
Solution said answer should be $27\pi/5$. I have no idea why. What am I doing wrong?
Yes, the answer that you were given is correct, since\begin{align}\int_0^3\left(\frac{y^2}3\right)^2\,\mathrm dy&=\frac19\int_0^3y^4\,\mathrm dy\\&=\frac19\left[\frac{y^5}5\right]_{y=0}^{y=3}\\&=\frac{27}5.\end{align}