Question

The volume of the solid obtained by revolving the region between the graphs of $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=x^3$ on $\left[0,1\right]$ around (a) the y-axis and (b) the line $x=2$.

Short answer

(a) The volume is $\frac{2}{5}\mathrm{\pi }$cubic units.

(b) The volume is$\frac{19}{15}\mathrm{\pi }$cubic units.

Step-by-step solution

Volume around y-axis

The required region is shown below,

$$\begin{gathered} y=\sqrt{x}\Rightarrow x=y^2 \\ y=x^3\Rightarrow x=y^{1/3} \end{gathered}$$

The volume is computed as,

$$\begin{gathered} V=\mathrm{\pi }{\int }_0^1\left[{\mathrm{y}}^{2/3}-{\mathrm{y}}^4\right]\mathrm{dy} \\ =\mathrm{\pi }{\left[\frac{3}{5}{\mathrm{y}}^{5/3}-\frac{{\mathrm{y}}^5}{5}\right]}_0^1 \\ =\mathrm{\pi }\left[\frac{3}{5}-\frac{1}{5}\right] \\ =\frac{2}{5}\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$

Volume around line x=2

The outer radius is $2-y^2$.

Inner radius is $2-y^{1/3}$.

The volume is computed as,

$$\begin{gathered} V=\mathrm{\pi }{\int }_0^1\left[{(2-{\mathrm{y}}^2)}^2-{(2-{\mathrm{y}}^{1/3})}^2\right]\mathrm{dy} \\ =\mathrm{\pi }{\int }_0^1\left[{\mathrm{y}}^4-4{\mathrm{y}}^2-{\mathrm{y}}^{2/3}+4{\mathrm{y}}^{1/3}\right]\mathrm{dy} \\ =\mathrm{\pi }{\left[\frac{{\mathrm{y}}^5}{5}-\frac{4}{3}{\mathrm{y}}^3-\frac{3}{5}{\mathrm{y}}^{5/3}+3{\mathrm{y}}^{4/3}\right]}_0^1 \\ =\mathrm{\pi }\left[\frac{1}{5}-\frac{4}{3}-\frac{3}{5}+3\right] \\ =\frac{19}{15}\mathrm{\pi }\mathrm{cubic}\mathrm{units}. \end{gathered}$$