Question

Consider the region between $f\left(x\right)=4-x^2$and the x-axis on $[0,2]$. For each line of rotation given in Exericses 29–32, use the shell method to construct definite integrals to find the volume of the resulting solid.

Short answer

The volume is $\frac{{\displaystyle 128}}{{\displaystyle 5}}\pi$.

Step-by-step solution

Step 1. Given Information. 

We are given,

Step 2. Finding the Volume. 

As the region bounded by$f\left(x\right)=4-x^2$ and the x-axis from $x=0$, to $x=2$is rotated around the line $y=4$, so to find the volume by shell, shells of height given by $x=f^{-1}\left(y\right)$ are drawn on the y-axis with average radius of $4-y$

To find $f^{-1}\left(y\right)$solve $y=4-x^2$ for x gives

data-custom-editor="chemistry" $x=\sqrt{4-y}$

So$f^{-1}\left(y\right)=\sqrt{4-y}$

Therefore using the shell method ,.

role="math" $$\begin{gathered} \text{Volume}=2\pi {\int }_0^4(4-y)\sqrt{4-y}dy \\ =2\pi {\left[-\frac{{\displaystyle (4-y{)}^{\frac{{\displaystyle 3}}{{\displaystyle 2}}+1}}}{{\displaystyle \frac{{\displaystyle 3}}{{\displaystyle 2}}+1}}\right]}_0^4 \\ =\frac{{\displaystyle 4\pi }}{{\displaystyle 5}}\left[-\left(0\right)+(4{)}^{\frac{{\displaystyle 5}}{{\displaystyle 2}}}\right] \\ =\frac{{\displaystyle 128}}{{\displaystyle 5}}\pi \end{gathered}$$

Hence, the volume is $\frac{{\displaystyle 128}}{{\displaystyle 5}}\pi$.