Q. 9 - Calculus

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 

Step-by-Step Solution

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 1: 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,

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

Step 2: Volume around the y - axis.

The volume is computed as,

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