Question

Consider the region between the graphs of $f\left(x\right)=5-x$and $g\left(x\right)=2x$on $\left[1,4\right]$For each line of rotation given in Exercises 45 and 46, use definite integrals to find the volume of the resulting solid.

Short answer

The volume of the solid is$\frac{1894}{27}\mathrm{\pi }$

Step-by-step solution

Step 1. Given Information

The given figure is

$$\begin{gathered} f\left(x\right)=g\left(x\right) \\ 5-x=2x \\ x=\frac{5}{3} \end{gathered}$$

Step 2. Finding Volume

$$\begin{gathered} V_1=\mathrm{\pi }{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(5-1\right)}^2-{\left(2\mathrm{x}-1\right)}^2\mathrm{dx} \\ {V}_1=\mathrm{\pi }{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(16-4{\mathrm{x}}^2-1+4\mathrm{x}\right)\mathrm{dx}=\mathrm{\pi }{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(15-4\mathrm{x}+3{\mathrm{x}}^2\right)\mathrm{dx} \\ {\mathrm{V}}_1=\mathrm{\pi }{\left[15\mathrm{x}-2{\mathrm{x}}^2+{\mathrm{x}}^3\right]}_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ =\mathrm{\pi }\left[\frac{75}{3}-\frac{50}{9}+\frac{125}{27}-\left(15-2+1\right)\right] \\ =\mathrm{\pi }\left(\frac{552}{27}-14\right)=\frac{174}{27}\mathrm{\pi }=\frac{58}{9}\mathrm{\pi } \end{gathered}$$

$$\begin{gathered} V_2=\mathrm{\pi }{\int }_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^4\left(3{\mathrm{x}}^2+4\mathrm{x}-15\right)\mathrm{dx} \\ {\mathrm{V}}_2=\mathrm{\pi }{\left[{\mathrm{x}}^3+2{\mathrm{x}}^2-15\mathrm{x}\right]}_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$/3$}\right.}^4 \\ =\mathrm{\pi }\left[64+32-60-\left(\frac{125}{27}+\frac{50}{9}-\frac{75}{3}\right)\right] \\ =\mathrm{\pi }\left(36+\frac{400}{27}\right)=\frac{1372}{27}\mathrm{\pi } \end{gathered}$$

Step 3. Finding Final Volume

$$\begin{gathered} V=V_1+V_2 \\ V=\frac{58}{9}\mathrm{\pi }+\frac{1372}{27}\mathrm{\pi }=\frac{1894}{27}\mathrm{\pi } \end{gathered}$$