Q:

Surface area of a revolution integral problem

$y=x^3, 0 \leq x \leq 2 $ rotated around the x-axis.
Here is my work,
$$\begin{align}
S &= \int_{0}^{2} 2\pi x^3 \sqrt{1 + (3x^2)^2}dx \\
&= 2\pi x^3 \sqrt{1+9x^4}dx \\
&= 2\pi \int_{0}^{2} \frac{1}{36}\sqrt{u}{du} \\
&= \frac{1}{18}\pi \int_{1}^{145} \sqrt{u} du \\
&= \frac{1}{18}\pi \bigg[\frac{u^{\frac{3}{2}}}{\frac{2}{3}} \bigg]***\\
&= \frac{1}{12}\pi \bigg[ 145\sqrt{145} - 1 \bigg]
\end{align}$$
However, the answer is actually $\frac{1}{27}\pi \bigg[ 145\sqrt{145} - 1 \bigg]$...I think I multipled the second to last step (labeled ***) wrong...but I thought when the fraction is in the denominator, you have to flip it so it becomes $\frac{1}{18} * \frac{3}{2}$ and not $\frac{1}{18} * \frac{2}{3}$?

A:

The step marked *** should actually read $\frac{u^{\frac{3}{2}}}{\frac{3}{2}}$.

