Question

The volume of the solid generated by revolving about the $y$ axis bounded by the parabola $y=x^2$and $x=y^2$ is

• A. $\frac{2}{15}\mathrm{\pi }$
• B. $\frac{24}{5}\mathrm{\pi }$
• C. $\frac{3}{10}\mathrm{\pi }$
• D. $\frac{5}{24}\mathrm{\pi }$

Answer 1

The correct option is C

$\frac{3}{10}\mathrm{\pi }$

Explanation of the correct option:

Given: Two curves $y=x^2$and $x=y^2$.

Let's plot the curves.

The shaded region is the region that is revolved.

$$\begin{gathered} V={\int }_0^1{\mathrm{\pi x}}^2\mathrm{dy} \\ \Rightarrow \mathrm{V}=\left|\mathrm{\pi }{\int }_0^1\left({\mathrm{y}}^4-\mathrm{y}\right)\mathrm{dy}\right| \\ \Rightarrow \mathrm{V}=\left|\mathrm{\pi }\left[\frac{{\mathrm{y}}^5}{5}-\frac{{\mathrm{y}}^2}{2}\right]\underset{0}{\overset{1}{}}\right| \\ \Rightarrow \mathrm{V}=\left|\mathrm{\pi }\left[\frac{1}{5}-\frac{1}{2}\right]\right| \\ \Rightarrow \mathrm{V}=\left|-\frac{3\mathrm{\pi }}{10}\right| \\ \Rightarrow \mathrm{V}=\frac{3\mathrm{\pi }}{10} \end{gathered}$$

Therefore, the volume of the solid generated by revolving the shaded area about the $y$ axis is $\frac{3\mathrm{\pi }}{10}$.

Hence option A is the correct option.