Q. 42 - Calculus

Question

Consider the region between the graph of $f\left(x\right)=4-x^2$and the line $y=5$ on $\left[0,2\right]$. For each line of rotation given in Exercises 41–44, use definite integrals to find the volume of the resulting solid.

Step-by-Step Solution

Answer

The volume of the solid is$\frac{20}{3}\mathrm{\pi }$

Step 1. Given Information

The given figure is

$$\begin{gathered} f\left(x\right)=4-x^2=y \\ x=\sqrt{4-y}=g\left(y\right) \end{gathered}$$

Step 2. Finding Volume

$V_1=\mathrm{\pi }{\int }_0^4{\left(2-\sqrt{4-\mathrm{y}}\right)}^2\mathrm{dy}=\mathrm{\pi }{\int }_0^4\left(8-\mathrm{y}-4\sqrt{4-\mathrm{y}}\right)\mathrm{dy}$

$$\begin{gathered} V_1=\mathrm{\pi }{\left[8\mathrm{y}-\frac{{\mathrm{y}}^2}{2}+\frac{8}{3}{\left(4-\mathrm{y}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^4 \\ {\mathrm{V}}_1=\mathrm{\pi }\left[32-8+0-\left(\frac{64}{3}\right)\right] \\ =\mathrm{\pi }\left(24-\frac{64}{3}\right)=\frac{8}{3}\mathrm{\pi } \end{gathered}$$

$$\begin{gathered} V_2=\mathrm{\pi }{\int }_4^{5}4\mathrm{dy}=4\mathrm{\pi }{\int }_4^{5}1\mathrm{dy} \\ =4\mathrm{\pi }\left(5-4\right)=4\mathrm{\pi } \end{gathered}$$

Now, $V=V_1+V_2$

$V=\frac{8}{3}\mathrm{\pi }+4\mathrm{\pi }=\frac{20}{3}\mathrm{\pi }$