Hello martin,
First, let's determine the coordinates of the intersections of the given curves. Equating them we find:
$$x^3=x^2$$
$$x^3-x^2=0$$
$$x^2(x-1)=0$$
Thus, we find the two points are $(0,0)$ and $(1,1)$. Here is a diagram of the region $T$:
View attachment 1933
a) Use the washer method to find the volume of the solid of revolution formed when $T$ is revolved about the $y$-axis.
The volume of an arbitrary washer is:
$$dV=\pi\left(R^2-r^2 \right)\,dy$$
where:
$$R=y^{\frac{1}{3}}$$
$$r=y^{\frac{1}{2}}$$
And so we have:
$$dV=\pi\left(y^{\frac{2}{3}}-y \right)\,dy$$
Summing up all the washers, there results:
$$V=\pi\int_0^1 y^{\frac{2}{3}}-y\,dy$$
Applying the FTOC, we get:
$$V=\pi\left[\frac{3}{5}y^{\frac{5}{3}}-\frac{1}{2}y^2 \right]_0^1=\pi\left(\frac{3}{5}-\frac{1}{2} \right)=\frac{\pi}{10}$$
b) Use the shell method to compute the volume of the solid of revolution described in part a).
The volume of an arbitrary shell is:
$$dV=2\pi rh\,dx$$
where:
$$r=x$$
$$h=x^2-x^3$$
And so we have:
$$dV=2\pi\left(x^3-x^4 \right)\,dx$$
Summing up all of the shells, there results:
$$V=2\pi\int_0^1 x^3-x^4\,dx$$
Applying the FTOC, we obtain:
$$V=2\pi\left[\frac{1}{4}x^3-\frac{1}{5}x^5 \right]_0^1=2\pi\left(\frac{1}{4}-\frac{1}{5} \right)=\frac{\pi}{10}$$
