Use both the Shell and Disk Methods to calculate the volume of the solid obtained by rotating the region under the graph of $f (x) = 2 - x^3$ for $0 \le x \le {2}^{\frac{1}{3}}$ about the $x$-axis and the $y$-axis.

Using the disk method, the volume $D_x$ of the solid obtained by rotating the region about the $x$-axis is $\int_0^a g(x)dx$ (this is the initial integral when you setup the problem), where
a =
g(x)=
$D_x$=

Using the shell method, the volume $S_x$ of the solid obtained by rotating the region about the $x$-axis is $\int_0^b h(y)dy$ (this is the initial integral when you setup the problem), where
b=
h(y)=
$S_x$=

Using the disk method, the volume $D_y$ of the solid obtained by rotating the region about the $y$-axis is $\int_0^A G(y)dy$ (this is the initial integral when you setup the problem), where
A =
G(y)=
$D_y$=

Using the shell method, the volume $S_y$ of the solid obtained by rotating the region about the $y$-axis is $\int_0^B H(x)dx$ (this is the initial integral when you setup the problem), where
B=
H(x)=
$S_y$=