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