Shell integration to find volume Find the volume generated by rotating the region bounded by the given curves about the y-axis.
$$y = \sqrt[3]{x}, y = 0, x = 1$$
Here is my work:
$$\begin{align}
V &= \int_{0}^{1}2\pi(x)\sqrt[3]{x}dx = 2\pi\int_{0}^{1}x^\frac{4}{3}dx \\
&= 2\pi \bigg[\frac{3}{6}x^\frac{6}{3} \bigg]_{0}^{1} = 2\pi\bigg[ \frac{1}{2}x^\frac{1}{2} \bigg]_{0}^{1}\\
&= \pi
\end{align}$$
However the answer is $\frac{6\pi}{7}$. It looks like you have to do integration by parts to integrate to get the answer. Why can't I "simplify" $x$ and $\sqrt[3]{x}$ by multiplying them together and adding the exponents? That seems to be reasonable in my mind, yet the result isn't the same

The setup is correct. The only thing that you got wrong is the antiderivative of the function $x^{\frac{4}{3}}$:
$$\int x^{\frac{4}{3}}\,dx=\frac{1}{\frac{4}{3}+1}x^{\frac{4}{3}+1}+C=
\frac{1}{\frac{7}{3}}x^{\frac{7}{3}}+C=
\frac{3x^{\frac{7}{3}}}{7}+C.$$
$$
V=2\pi\int_{0}^{1}x\sqrt[3]{x}\,dx=
2\pi\int_{0}^{1}x^{\frac{4}{3}}\,dx=
2\pi\left[\frac{3}{7}x^{\frac{7}{3}}\right]_{0}^{1}=\\
2\pi\left(\frac{3}{7}1^{\frac{7}{3}}-\frac{3}{7}0^{\frac{7}{3}}\right)=
2\pi\left(\frac{3}{7}-0\right)=\frac{6\pi}{7}\ cubic\ units.
$$
