Chapter 7.3, Problem 38E

To determine

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

Answer to Problem 38E

The volume of the solid generated is $V=\frac{5\pi }{6}\text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

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

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 $y=x^2$ and the lines $x=0\text{ and }y=2-x$ about the $y$ -axis.

Taking into account integration in terms of $y$ since the specified region revolves around the $y$ -axis. The illustration suggests that the region must divide into two parts: a region bounded by the lines $y=1,x=0$ and the line $y=2-x$ , and a region bounded by the curve $y=x^2$ and the lines $y=1\text{ and }x=0$ .

Now to find the limits of integration $y_1=a,y_2=y_3=b\text{ and }{y}_4=c$ .

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

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

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

From the graph the limits will be the $y$ -coordinates of intersection of the curve $y=x^2$ and the lines $x=0\text{ and }y=2-x$ .

Therefore, the limits are $a=0,b=1\text{ and }c=2$ .

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

Rewriting the equation,

  $\begin{array}{c}y=x^2\\ x=\sqrt{y}\end{array}$

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

The area of cross section is given by,

  $A_1\left(y\right)\text{=y}\pi$

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

Rewriting the equation,

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

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

The area of cross section is given by,

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

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

  $\begin{array}{c}{V}_1={\displaystyle \underset{0}{\overset{1}{\int }}{A}_1\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}y\pi dy}\\ ={\left.\frac{y^2\pi }{2}\right|}_0^1\\ =\frac{\pi }{2}\\ V_2={\displaystyle \underset{1}{\overset{2}{\int }}{A}_2\left(y\right)dy}\\ ={\displaystyle \underset{1}{\overset{2}{\int }}\left(4-4y+y^2\right)\pi dy}\\ ={\left.\left(4y-2y^2+\frac{y^3}{3}\right)\pi \right|}_1^2\\ =\pi \left(8-8+\frac{8}{3}-4+2-\frac{1}{3}\right)\\ =\left(-2+\frac{7}{3}\right)\pi \\ =\frac{\pi }{3}\end{array}$

The volume of the solid is given by,

  $\begin{array}{c}V=V_1+V_2\\ =\frac{\pi }{3}+\frac{\pi }{2}\\ V=\frac{5\pi }{6}\end{array}$

Therefore, the volume of the solid generated is $V=\frac{5\pi }{6}\text{ cu}\text{. units}$ .