Chapter 7.3, Problem 40E

To determine

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

Answer to Problem 40E

The volume of the solid generated is $V=\frac{7}{15}\pi \text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=2x-1\\ y=\sqrt{x}\\ x=0\end{array}$

Calculation:

Let's graph this area in Desmos and draw some conclusions based on it.

  

From the graph, the region is bounded by the curve $y=\sqrt{x}$ and the lines $x=4\text{ and }y=2x-1$ about the $y$ -axis.

Let’s consider the cases $y\in \left[-1,0\right]\text{ and }y\in \left[0,1\right]$ .

Considering the case $y\in \left[-1,0\right]$ the region bounded by the curve $x=y^2$ and the lines $x=\frac{y}{2}+\frac{1}{2}\text{ and }y=0$ .

Considering the case $y\in \left[0,1\right]$ the region bounded by the curve $x=y^2$ and lines $x=0,x=\frac{y}{2}+\frac{1}{2}\text{ and }y=0$ .

The limits of integration $a=0\text{ and }b=1$ .

Finding the formulas for $A_1\left(y\right)\text{ and }{A}_2\left(y\right)$ on height $y\in \left[0,1\right]\text{ and }y\in \left[-1,0\right]$ perpendicular to the $y$ -axis,

  $\begin{array}{c}{V}_1={\displaystyle \underset{0}{\overset{1}{\int }}{A}_1\left(y\right)dy}\\ V_2={\displaystyle \underset{-1}{\overset{0}{\int }}{A}_2\left(y\right)dy}\end{array}$

Thus, the volume of the solid is the sum of these integrals.

To find the formula for $A_1\left(y\right)$ ,

Since each cross section on height $y\in \left[0,1\right]$ perpendicular to the $y$ -axis is a circular region of radius $\frac{y}{2}+\frac{1}{2}$ .

The area of cross section is given by,

  $\begin{array}{c}{A}_1\left(y\right)\text{=}\left({\left(\frac{y}{2}+\frac{1}{2}\right)}^2-y^4\right)\pi \\ =\left(-y^4+\frac{y^4}{4}+\frac{y}{2}+\frac{1}{4}\right)\pi \end{array}$

To find the formula for $A_2\left(y\right)$ ,

Since each cross section on height $y\in \left[-1,0\right]$ perpendicular to the $y$ -axis is a circular region of radius $\frac{y}{2}+\frac{1}{2}$ .

The area of cross section is given by,

  $\begin{array}{c}{A}_2\left(y\right)\text{=}{\left(\frac{y}{2}+\frac{1}{2}\right)}^2\pi \\ =\left(\frac{y^2}{4}+\frac{y}{2}+\frac{1}{4}\right)\pi \end{array}$

Finding the values of $V_1\text{ and }{V}_2$ ,

  $\begin{array}{c}{V}_1={\displaystyle \underset{0}{\overset{1}{\int }}{A}_1\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{\left(-y^4+\frac{y^2}{4}+\frac{y}{2}+\frac{1}{4}\right)}^2\pi dy}\\ ={\left.\left(-\frac{y^5}{5}+\frac{y^3}{12}+\frac{y^2}{4}+\frac{y}{4}\right)\pi \right|}_0^1\\ =-\frac{1}{5}+\frac{1}{12}+\frac{1}{4}+\frac{1}{4}\\ =\frac{23}{60}\pi \\ V_2={\displaystyle \underset{-1}{\overset{0}{\int }}{A}_2\left(y\right)dy}\\ ={\displaystyle \underset{-1}{\overset{0}{\int }}{\left(\frac{y^2}{4}+\frac{y}{2}+\frac{1}{4}\right)}^2\pi dy}\\ ={\left.\left(\frac{y^3}{12}+\frac{y^2}{4}+\frac{y}{4}\right)\pi \right|}_{-1}^0\\ =\frac{1}{12}-\frac{1}{4}+\frac{1}{4}\\ =\frac{5}{60}\pi \end{array}$

The volume of the solid is given by,

  $\begin{array}{c}V=V_1+V_2\\ =\frac{23}{60}\pi +\frac{5}{60}\pi \\ V=\frac{7}{15}\pi \end{array}$

Therefore, the volume of the solid generated is $V=\frac{7}{15}\pi \text{ cu}\text{. units}$ .