Q. 30 - Calculus

Question

Think about the area between the x-axis on $[0, 2]$ and $f\left(x\right)=4-x^2$. Use the shell approach to create definite integrals for each line of rotation provided in this exercise to determine the volume of the resulting solid. Around the x-axis.

Step-by-Step Solution

Answer

The volume of the solid is $\frac{512}{15}\pi \approx 34.33\pi$.

Step 1: Given information

Consider the given information,

$f\left(x\right)=4-x^2$

Step 2: Calculation

To find the volume using the shell method, shells of height determined by $f^{-1}\left(y\right)$   are drawn on the y-axis with the average radius of y. This is done because the region bounded by $f\left(x\right)=4-x^2$and the x-axis from $x=0$ to $x=2$ are rotated around the y-axis.

To find $f^{-1}\left(y\right)$solve $y=4-x^2$ for x gives $x=\sqrt{4-y}$. So $f^{-1}\left(y\right)=\sqrt{4-y}$  Consequently, using the shell method, $\mathrm{Volume}=2\pi {\int }_0^4\left(y\right)\sqrt{4-y}dy.$

To solve the integral let $4-y=t^2,$ then $dy=-2tdt$ and when$y=0, t=2$ and when $y=4, t=0$

Consequently.

$\begin{array}{rcl}\mathrm{Volume}& =& 2\pi {\int }_2^0\left(4-t^2\right)\sqrt{t^2}(-2tdt)\\ & =& 4\pi {\int }_0^2\left(4t^2-t^4\right)dt \mathrm{changing} \mathrm{the} \mathrm{limits} \mathrm{of} \mathrm{integration}\\ & =& 4\pi {\left[4\frac{t^3}{3}-\frac{t^5}{5}\right]}_0^2 \mathrm{up} \mathrm{on} \mathrm{integration}\\ & =& 4\pi \left(\left[\frac{32}{3}-\frac{32}{5}\right]-[0-0]\right) \mathrm{simplifying}\end{array}$

Hence, the volume of the solid is $\frac{512}{15}\pi \approx 34.33\pi$