Chapter 8.3, Problem 42E

To determine

To Find : The volume of the given solid.

Answer to Problem 42E

The required volume is $\frac{8}{3}$ Unit³.

Explanation of Solution

Given information : The base of the solid disk $x^2+y^2\le 1$ .

  

The cross-section by planes perpendicular to Y-axis between $y=-1$ and $y=1$ are isosceles right triangle with one leg in the disk.

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 isosceles right triangle of equal sides “ $a$ ” is $\frac{1}{2}{a}^2$ Unit².

Calculation :

The cross-section by plane perpendicular to the Y-axis are isosceles right triangle.

The length of the leg on the disc is $2x$ .

So the area of that isosceles right triangle is $A\left(y\right)=\frac{1}{2}{\left(2x\right)}^2$

Now,

  $\begin{array}{l}{x}^2+y^2=1\\ \Rightarrow x^2=1-y^2\end{array}$

  $\therefore A\left(y\right)=\frac{1}{2}\times 4\left(1-y^2\right)=2\left(1-y^2\right)$

So the volume of the solid is,

  $\begin{array}{l}{\displaystyle \underset{-1}{\overset{1}{\int }}A\left(y\right)}dy\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}2\left(1-y^2\right)dy}\\ =2{\displaystyle \underset{-1}{\overset{1}{\int }}dy}-2{\displaystyle \underset{-1}{\overset{1}{\int }}{y}^2}dy\\ =2{\left[y\right]}_{-1}^1-2{\left[\frac{y^3}{3}\right]}_{-1}^1\\ =2\left[1-\left(-1\right)\right]-2\left[\frac{1}{3}-\left(-\frac{1}{3}\right)\right]\\ =4-\frac{4}{3}\\ =\frac{8}{3}\end{array}$

Therefore, the required volume is $\frac{8}{3}$ Unit³.