## Calculus (3rd Edition)

$$\frac{3\pi}{10}$$
Given $$x=y^2,\ \ x=\sqrt{y},$$ We first find the intersection points \begin{align*} y^2&=\sqrt{y}\\ y^2-\sqrt{y}&=0\\ \sqrt{y}(y^{3/2}-1)&=0 \end{align*} Then $y=0,\ 1$ Then the volume of the solid obtained by rotating the region enclosed by the graph of $x=y^2,\ \ x=\sqrt{y}$ about the $y-$axis is given by \begin{align*} V&=\pi\int_{0}^{1}[f(y)]^2-[g(y)]^2dy\\ &= \pi \int_{0}^{1} [ y-y^4]dy\\ &=\pi \left( \frac{1}{2}y^2- \frac{1}{5}y^5\right)\bigg|_{0}^{1}\\ &=\frac{3\pi}{10} \end{align*} We use the Mathematica program to plot the solid.