Volume with a Cubic Find the volume of the solid of revolution formed when the curve $y=x^3+1$ is rotated around the $y$-axis from $y=1$ to $y=9$.
I understand the rule to find the volume when it is rotated around the $y$-axis; however, how do I make the subject and squared when it is cubed?
Does that make sense?
 A: Make cylindrical shells of radius $x$ $\>(0\leq x\leq2)$, height $9-(x^3+1)=8-x^3$, and thickness $dx$. The volume $V$ in question then appears as
$$V=\int_0^2 2\pi x(8-x^3)\>dx=2\pi\left(4x^2-{x^5\over5}\right)\Biggr|_0^2={96\over5}\pi\ .$$
A: We're working on the $y$ axis, so let's change orientations first. 
$$x = (y-1)^{1/3}$$
Since we're asked to find the volume of revolution, we need to determine the radius and area of the revolution at each point along the given interval $y \in [1, 9]$. The radius is simply given by the "height" of the function $x$, so the area is given by $\pi x^2 = \pi (y-1)^{2/3}$. Integrating along the interval, we find
$$ \int_{y=1}^9 \pi (y-1)^{2/3} \,\mathrm{d}x = \left.\frac{3\pi}{5}(y-1)^{5/3}\right|_{1}^{9} = \frac{96\pi}{5} $$
A: Since $y=x^3+1$, we have $\mathrm{d}y=3x^2\mathrm{d}x$. Therefore, using the Disc Method for finding the volume of rotation, we get
$$
\begin{align}
\int_1^9\pi x^2\,\mathrm{d}y
&=\int_0^2\pi x^2\,3x^2\,\mathrm{d}x\\
&=\left[3\pi\frac{x^5}5\right]_0^2\\[4pt]
&=\frac{96\pi}5
\end{align}
$$
A: hint: $V = \displaystyle \int_{1}^9 \pi(\sqrt[3]{y-1})^2dy$
