Question

Consider the region between the graph of 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.

Short answer

The volume of the solid is$\frac{206}{15}\mathrm{\pi }$

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

$$\begin{gathered} V=\mathrm{\pi }{\int }_a^b{\left(\mathrm{R}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}={\int }_0^2{\left(5-\left(4-{\mathrm{x}}^2\right)\right)}^2\mathrm{dx} \\ V={\int }_0^2\left({\mathrm{x}}^4+2{\mathrm{x}}^2+1\right)\mathrm{dx} \end{gathered}$$

$$\begin{gathered} V=\mathrm{\pi }{{\left(\frac{x^5}{5}+\frac{2x^3}{3}+x\right)}^2}_0 \\ V=\mathrm{\pi }(\frac{32}{5}+\frac{16}{3}+2-0) \\ V=\frac{96+80+30}{15}=\frac{206}{15}\mathrm{\pi } \end{gathered}$$