Chapter 7.3, Problem 39E

To determine

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

Answer to Problem 39E

The volume of the solid generated is $V=\frac{128\pi }{5}\text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=\sqrt{x}\\ y=0\\ x=4\end{array}$

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 $y=0\text{ and }x=4$ about the $y$ -axis.

Taking into account integration in terms of $y$ since the specified region revolves around the $y$ -axis.

Now to find the limits of integration $y_1=a\text{ and }{y}_2=b$ .

Finding the formulas for $A\left(y\right)$ on height $y\in \left[a,b\right]$ perpendicular to the $y$ -axis,

  $V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(y\right)dy}$

From the graph the limits will be the $y$ -coordinates of intersection of the curve $y=\sqrt{x}$ and the lines $y=0$ and the $y$ -coordinates of intersection of the curve $y=\sqrt{x}$ and the lines $x=4$ .

Since the intersection of the curve $y=\sqrt{x}$ and the lines $y=0$ is a point equals to $0$ and the intersection of the curve $y=\sqrt{x}$ and the lines $x=4$ is a point equals to $y=2$ .

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

Rewriting the equation,

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

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

The area of cross section is given by,

  $\begin{array}{c}A\left(y\right)\text{=}\underset{\begin{array}{l}\text{ Area of}\\ \text{bigger circle}\end{array}}{\underbrace{{\text{4}}^2\pi }}-\underset{\begin{array}{l}\text{ Area of}\\ \text{smaller circle}\end{array}}{\underbrace{y^4\pi }}\\ =\left(16-y^4\right)\pi \end{array}$

Finding the volume of the solid,

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{2}{\int }}A\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\left(16-y^4\right)\pi dy}\\ ={\left.\pi \left(16y-\frac{y^5}{5}\right)\right|}_0^2\\ =\pi \left(32-\frac{32}{5}-\left(0-0\right)\right)\\ V=\frac{128\pi }{5}\text{ cu}\text{. units}\end{array}$

Therefore, the volume of the solid generated is $V=\frac{128\pi }{5}\text{ cu}\text{. unit}$ .