Chapter 8.3, Problem 41E

To determine

To find : The volume of the solid.

Answer to Problem 41E

The required volume of the solid is approximately $25.1327$ Unit³.

Explanation of Solution

Given information : The solid lies between the planes perpendicular to the Y-axis at $y=0$ and $y=2$ .

The cross-sections perpendicular to the Y-axis are circular disk with diametes running from Y-axis to the parabola $x=\sqrt{5}{y}^2$ .

Formula used :

The volume of a solid of known integrable cross-section area $A\left(y\right)$ from $y=a$ to $y=b$ is the integral of $A$ from $a$ to $b$ .

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

The area of a circular disk of diameter “ $d$ ” is $\pi {\left(\frac{d}{2}\right)}^2$ Unit².

Calculation :

The cross-section perpendicular to the Y-axis are circular disks.

So, the area of the circular disk is,

  $\begin{array}{c}A\left(y\right)=\pi {\left(\frac{x}{2}\right)}^2\\ =\frac{\pi }{4}{\left(\sqrt{5}{y}^2\right)}^2\\ =\frac{5\pi }{4}{y}^4\end{array}$

The Volume of the solid is ,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}A\left(y\right)}dy\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\frac{5\pi }{4}}{y}^{4}dy\\ =\frac{5\pi }{4}{\displaystyle \underset{0}{\overset{2}{\int }}{y}^{4}dy}\\ =\frac{5\pi }{4}{\left[\frac{y^5}{5}\right]}_0^2\\ =\frac{5\pi }{4}\left[\frac{2^5}{5}-0\right]\\ =\frac{5\pi }{4}\times \frac{2^5}{5}\\ =8\pi \\ \approx 25.1327\end{array}$

Therefore, required volume of the solid is approximately $25.1327$ Unit³.