Question

Consider the region between the graph of $f\left(x\right)=x-2$ and the x-axis on [2,5]. For each line of rotation given in Exercises 35– 40, use definite integrals to find the volume of the resulting solid.

Short answer

The required volume is$36\mathrm{\pi }$

Step-by-step solution

Step 1. Given Information     

The given figure is

Step 2. Calculation      

Express the curve as inverse function.

$$\begin{gathered} y=x-2 \\ x=y+2 \\ g\left(y\right)=y+2 \end{gathered}$$

For the x-interval of $[2,5]$, the corresponding interval of y-variable will be $[0,3]$

For the washer the external radius of each washer is$x=5$and internal radius of each wahser is given by$g\left(y\right)$

Step 3. Calculation

Determine the volume of solid of revolution.

$$\begin{gathered} V=\mathrm{\pi }{\int }_0^3\left(5^2-{\left(\mathrm{g}\left(\mathrm{y}\right)\right)}^2\right)\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^3\left(25-{\left(\mathrm{y}+2\right)}^2\right)\mathrm{dy} \\ =\mathrm{\pi }{\int }_0^3\left(25-{\mathrm{y}}^2-4\mathrm{y}-4\right)\mathrm{dy} \\ =\mathrm{\pi }{\int }_0^3\left(21-{\mathrm{y}}^2-4\mathrm{y}\right)\mathrm{dy} \\ =\mathrm{\pi }{\left[21\mathrm{y}-\frac{{\mathrm{y}}^3}{3}-\frac{4{\mathrm{y}}^2}{2}\right]}_0^3\mathrm{dy} \\ =\mathrm{\pi }\left(\left(21\left(3\right)-\frac{3^3}{3}-\frac{4{\left(3\right)}^2}{2}\right)-\left(21\left(0\right)-\frac{0^3}{3}-\frac{4{\left(0\right)}^2}{2}\right)\right) \\ =\mathrm{\pi }\left[\left(63-9-18\right)-0\right] \\ =36\mathrm{\pi } \end{gathered}$$