Question

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

Short answer

The volume of the solid is$16\mathrm{\pi }$

Step-by-step solution

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

$$\begin{gathered} V_1=\mathrm{\pi }{\int }_0^4{\left(3-\left(\sqrt{4-\mathrm{y}}\right)-1\right)}^2\mathrm{dy} \\ {\mathrm{V}}_1=\mathrm{\pi }{\int }_0^4\left(12-\mathrm{y}-6\sqrt{4-\mathrm{y}}\right)\mathrm{dy} \\ {V}_1=\mathrm{\pi }{\left[12\mathrm{y}-\frac{{\mathrm{y}}^2}{2}-6\left(\frac{{\left(4-\mathrm{y}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\right]}_0^4={\mathrm{V}}_1=\mathrm{\pi }{\left[12\mathrm{y}-\frac{{\mathrm{y}}^2}{2}+4{\left(4-\mathrm{y}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^4 \\ {\mathrm{V}}_1=\mathrm{\pi }\left[48-8+4\left(0\right)-\left(32\right)\right]=8\mathrm{\pi } \end{gathered}$$

and

$$\begin{gathered} V_2=\mathrm{\pi }{\int }_4^5\left(9-1\right)dy=8\mathrm{\pi }{\int }_4^{5}dy \\ {V}_2=8\mathrm{\pi }\left[5-4\right]=8\mathrm{\pi } \end{gathered}$$

Now,

$$\begin{gathered} V=V_1+V_2 \\ V=8\mathrm{\pi }+8\mathrm{\pi }=16\mathrm{\pi } \end{gathered}$$