Q. 10 - Calculus

Question

The volume of the solid obtained by revolving the region between the graph of $f\left(x\right)=9-x^2$ and the x-axis on $\left[-3, 3\right]$ around (a) the x-axis and (b) the line $y=-3$. 

Step-by-Step Solution

Answer

(a) The volume is $\frac{1296}{5}\mathrm{\pi }$ cubic units.

(b) The volume is $\frac{216}{5}\mathrm{\pi }$ cubic units.

Step 1: Volume around x - axis

The required region is shown below,

The volume is computed as,

$$\begin{gathered} V=\mathrm{\pi }{\int }_{-3}^3{\mathrm{y}}^2 \mathrm{dx} \\ =\mathrm{\pi }{\int }_{-3}^3{(9-{\mathrm{x}}^2)}^2 \mathrm{dx} =2\mathrm{\pi }{\int }_0^3\left(81+{\mathrm{x}}^4-18{\mathrm{x}}^2\right) \mathrm{dx} \\ =2\mathrm{\pi }{\left[81\mathrm{x}+\frac{{\mathrm{x}}^5}{5}-6{\mathrm{x}}^3\right]}_0^3 \\ =\frac{1296}{5}\mathrm{\pi } \mathrm{cubic} \mathrm{units} \end{gathered}$$

Step 2: Volume around y = - 3

The outer radius of the cross-section is,

$y=6-x^2$

The inner radius is, $y=-3$

The volume is computed as,

$$\begin{gathered} V=\mathrm{\pi }{\int }_{-3}^3\left[{(6-{\mathrm{x}}^2)}^2-{(-3)}^2\right] \mathrm{dx} \\ =2\mathrm{\pi }{\int }_0^3\left(36+{\mathrm{x}}^4-12{\mathrm{x}}^2-9\right) \mathrm{dx} \\ =2\mathrm{\pi }{\left[27\mathrm{x}+\frac{{\mathrm{x}}^5}{5}-4{\mathrm{x}}^3\right]}_0^3 \\ =\frac{216}{5}\mathrm{\pi } \mathrm{cubic} \mathrm{units} \end{gathered}$$