Chapter 8.3, Problem 6E

To determine

To find : the volume of the solid.

Answer to Problem 6E

The volume of the solid is $\frac{8}{3}$ .

Explanation of Solution

Given information : The solid lies between planes perpendicular to the x -axis at $x=-1\&x=1$ . The cross sections perpendiculars to the x-axis between these planes are squares whose diagonals run from the semicircle $y=-\sqrt{1-x^2}$ to the semicircle $y=\sqrt{1-x^2}$ .

Calculation :

The diagonal of the square cross section has length ${\left(2\sqrt{1-x^2}\right)}^2$ .

The area of the cross section is $A\left(x\right)={\left(2\sqrt{1-x^2}\right)}^2$ .

  $\begin{array}{l}V={\displaystyle \underset{-1}{\overset{1}{\int }}A\left(x\right)dx}\\ V={\displaystyle \underset{-1}{\overset{1}{\int }}{\left(2\sqrt{1-x^2}\right)}^{2}dx}\\ V=\frac{8}{3}\end{array}$