Q. 41 - 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 $15\mathrm{\pi }$

Step 1. Given Information

The given figure is

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

$x=\sqrt{4-y}=g\left(y\right)$

Step 2. Finding Volume

$$\begin{gathered} V_1=\mathrm{\pi }{\int }_0^4(2-{\left(\sqrt{4-\mathrm{y}}\right)}^2\mathrm{dy} \\ {V}_1=\mathrm{\pi }{\int }_0^4(8-\mathrm{y}-2\sqrt{4-\mathrm{y}})\mathrm{dy}=\mathrm{\pi }{\left[8y-\frac{y^2}{2}+2\frac{{\left(4-y\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3/2}\right]}_0^4 \\ {V}_1=11\mathrm{\pi } \end{gathered}$$

and 

$$\begin{gathered} V_2=\mathrm{\pi }{\int }_4^5{2}^2\mathrm{dy}=4\mathrm{\pi }{\int }_4^5\mathrm{dy} \\ {V}_2=4\mathrm{\pi }\left(5-4\right)=4\mathrm{\pi } \end{gathered}$$

Thus, Final volume is

 $$\begin{gathered} V=V_1+V_2 \\ V=11\mathrm{\pi }+4\mathrm{\pi }=15\mathrm{\pi } \end{gathered}$$