Q. 45 - Calculus

Question

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

Step-by-Step Solution

Answer

The volume of the solid is$\frac{1925}{27}\mathrm{\pi }$

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

$$\begin{gathered} V_1=\mathrm{\pi }{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(5-\mathrm{x}\right)}^2-{\left(2\mathrm{x}\right)}^2\mathrm{dx} \\ {V}_1=\mathrm{\pi }{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(25+{\mathrm{x}}^2-10\mathrm{x}-4{\mathrm{x}}^2\right)\mathrm{dx} \\ {\mathrm{V}}_1=\mathrm{\pi }{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(-3{\mathrm{x}}^2-10\mathrm{x}+25\right)\mathrm{dx} \\ {\mathrm{V}}_1=\mathrm{\pi }{\left[-{\mathrm{x}}^3-5{\mathrm{x}}^2+25\mathrm{x}\right]}_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ {\mathrm{V}}_1=\mathrm{\pi }\left[(-\frac{125}{27}-\frac{125\times 3}{9\times 3}+\frac{125\times 9}{3\times 9})-\left(-1-5+25\right)\right] \\ {\mathrm{V}}_1=\mathrm{\pi }\left[\frac{625}{27}-19\right]=4.14\mathrm{\pi } \end{gathered}$$

and

$$\begin{gathered} V_2=\mathrm{\pi }{\int }_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^4{\left(2\mathrm{x}\right)}^2-{\left(5-\mathrm{x}\right)}^2\mathrm{dx} \\ {V}_2=\mathrm{\pi }{\int }_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^4\left(4{\mathrm{x}}^2-25-{\mathrm{x}}^2+10\mathrm{x}\right)\mathrm{dx} \\ {\mathrm{V}}_2=\mathrm{\pi }{\int }_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^4\left(3{\mathrm{x}}^2+10\mathrm{x}-25\right)\mathrm{dx} \\ {\mathrm{V}}_2=\mathrm{\pi }{\left[{\mathrm{x}}^3+5{\mathrm{x}}^2-25\mathrm{x}\right]}_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^4 \\ {\mathrm{V}}_2=\mathrm{\pi }\left[\left(64+80-100\right)-\left(\frac{125}{27}+\frac{125\times 3}{9\times 3}-\frac{125\times 9}{3\times 9}\right)\right] \\ =\mathrm{\pi }\left[44+\frac{625}{27}\right]=67.14\mathrm{\pi } \end{gathered}$$

Step 3. Finding Final Volume

$$\begin{gathered} V=V_1+V_2 \\ V=4.14\mathrm{\pi }+67.14\mathrm{\pi }=71.29\mathrm{\pi } \end{gathered}$$