Answer
$\frac{16\pi}{35}$
Work Step by Step
Step 1. Draw a diagram as shown in the figure. The limits of integration with respect to $y$ can be found from $0$ to $1$.
Step 2. Using the washer approximation, we have
$dV=\pi (x_2^2-x_1^2)dy=\pi (y^{2/3}-y^6)dy$
Step 3. We have
$V=\int_0^1 \pi (y^{2/3}-y^6)dy=\pi (\frac{3}{5}y^{5/3}-\frac{1}{7}y^7)|_0^1=\frac{16\pi}{35}$
