A region in the first quadrant is bounded on the left by the $y$-axis, above by $y=4$ and below by the graph of
\begin{equation*} \ y = x^3 \end{equation*}
Find the volume of if this region is revolved about the $x = 3$.
My solution:
\begin{equation*} \int_{4^{1/3}}^{3}{\pi} {x^6} dx \end{equation*}
Does this look okay? Any help is highly appreciated.
Thanks
Integrating with respect to $y$ seems easier, if only because the intersection of $y=x^3$ and $y=4$ has a not-nice $x$-coordinate ($\sqrt[3]{4}$).
We're going to take the volume of the big cylinder with radius $3$ and height $4$, and then subtract the volume of the solid whose cross sections in $y$ are circles with radius $3-\sqrt[3]{y}$. This is
$$V=36\pi - \int_0^4 \pi\left(3-\sqrt[3]{y}\right)^2 \,\mathrm{d}y$$
Expanding the squared term gives
$$V=36\pi - \left(\int_0^4 9 \,\mathrm{d}y-\int_0^4 6y^{1/3}\,\mathrm{d}y + \int_0^4 y^{2/3} \,\mathrm{d}y\right)\pi$$
$$V=\pi\left(36-9\int_0^4 1 \,\mathrm{d}y + 6\int_0^4 y^{1/3}\,\mathrm{d}y-\int_0^4 y^{2/3}\,\mathrm{d}y\right)$$
These are now just standard integrals. Finish the beast!