<s>[INST] Find the volume of a bounded region rotated about the $x$-axis Question
A region is bounded by the $x$-axis, $y=x^2-9$ and $y=x^2-4$, where the region looks similar to a u-shape. The region is then rotated about the $x$-axis. Find the volume of the solid formed.
My Working
We can see that $y=x^2-9$ intersects the $x$-axis at $-3$ and $3$, while $y=x^2-4$ intersects the $x$-axis at $-2$ and $2$. The volume should be
\begin{align}
V&=\pi\left[\int^3_{-3}(x^2-9)^2dx-\int^2_{-2}(x^2-4)^2dx\right]\\
&=2\pi\left[\int^3_0 (x^4-18x^2+81) dx-\int^2_0(x^4-8x^2+16) dx\right]\\
&=2\pi\left\{\left[\frac{x^5}{5}-6x^3+81x\right]^3_0-\left[\frac{x^5}{5}-\frac{8x^3}{3}+16x\right]^2_0\right\}\\
&=2\pi\left(\frac{243}{5}-162+243-\frac{32}{5}+\frac{64}{3}-32\right)\\
&=2\pi\left(49+\frac{209}{5}+\frac{64}{3}\right)\\
&=2\pi\cdot\frac{735+627+320}{15}\\
&=2\pi\cdot\frac{1682}{15}=\frac{3364}{15}\pi
\end{align}
But my teacher said the answer was $$\frac{3376}{15}\pi$$
Could anyone please point out where I got the answer wrong? Thank you!
 [/INST] After looking at CyclotomicField's comment, it seems that the teacher's answer is right and that I have just made a silly mistake, as shown below:
\begin{align}
V&=2\pi\left(49+\frac{211}{5}+\frac{64}{3}\right)\\
&=2\pi\cdot\frac{735+633+320}{15}\\
&=2\pi\cdot\frac{1688}{15}=\frac{3376}{15}\pi
\end{align}
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