Chapter 7, Problem 44RE

To determine

To find:The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 44RE

The volume of the solid generated is $V=\pi \text{ cu}\text{. unit}$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=2x\\ y=\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ x=1\end{array}$

Formula used:

The formula for cylindrical shell is,

  $V={\displaystyle {\int }_a^{b}2\pi xf\left(x\right)dx}$

Where,

  $a$ = left boundary of the area on the $x$ -axis

  $b$ = right boundary of the area on the $y$ -axis

Calculation:

Let's graph thegiven area and draw some conclusions based on it.

  

When two functions encircle a region, the difference between the two functions stands in for the original function is,

  $V={\displaystyle {\int }_a^{b}2\pi x}\left(f\left(x\right)-g\left(x\right)\right)dx$

It is known that,

  $\begin{array}{c}f\left(x\right)=2x\\ g\left(x\right)=\frac{x}{2}\\ a=0\\ b=1\end{array}$

Substituting the values in the formula to get,

  $\begin{array}{c}V={\displaystyle {\int }_a^{b}2\pi x}\left(f\left(x\right)-g\left(x\right)\right)dx\\ ={\displaystyle {\int }_0^{1}2\pi x\left(2x-\frac{x}{2}\right)dx}\\ ={\displaystyle {\int }_0^{1}2\pi x\left(\frac{3x}{2}\right)dx}\\ =3\pi {\displaystyle {\int }_0^1{x}^{2}dx}\\ =3\pi {\left.\left(\frac{x^3}{3}\right)\right|}_0^1\\ =3\pi \left(\frac{1}{3}-\frac{0}{3}\right)\\ =3\pi \frac{1}{3}\\ V=\pi \end{array}$

Therefore, the volume of the solid generated is $V=\pi \text{ cu}\text{. unit}$ .