Question

The volume of the solid obtained by revolving the region between the graph of $f\left(x\right)=x^2$and the $y-$axis for $0\le x\le 2$around $\left(a\right)$the $x$-axis and $\left(b\right)$ the $y$-axis

Short answer

(a) The volume of solid about $x$- axis is $\frac{32}{5}\mathrm{\pi }$cubic units.

(b) The volume of solid about $y$-axis is $8\mathrm{\pi }$cubic units.

Step-by-step solution

Volume around x-axis.

The graph of $f\left(x\right)=x^2,0\le x\le 2$is shown below,

The volume is computed as,

$$\begin{gathered} V={\int }_a^b{\mathrm{\pi y}}^2\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^2{\left({\mathrm{x}}^2\right)}^2\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^2{\mathrm{x}}^4\mathrm{dx} \\ =\mathrm{\pi }{\left[\frac{{\mathrm{x}}^5}{5}\right]}_0^2 \\ =\frac{32}{5}\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$

Volume around the y- axis.

The volume is computed as,

$$\begin{gathered} V={\int }_a^b{\mathrm{\pi x}}^2\mathrm{dy} \\ =\mathrm{\pi }{\int }_0^4\mathrm{y}\mathrm{dy} \\ =\mathrm{\pi }{\left[\frac{{\mathrm{y}}^2}{2}\right]}_0^4 \\ =\frac{16}{2}\mathrm{\pi } \\ =8\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$