Chapter 7.3, Problem 44E

To determine

To find:The volume of the given solid.

Answer to Problem 44E

The volume of the solid generated is $V=\frac{8}{3}\text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

Thebase of the solid is given by $x^2+y^2\le 1$ .

Calculation:

  

Let’s consider the integration in terms of $y$ since cross-section perpendicular to $y$ -axisand they are isosceles right triangles.

In order to get the cathetus of the triangle's cross-section on high $y=a$ , we must determine the length of the segment that results from the junction of the $y=a$ line and the circle.

The limits of integration are $-1\text{ and 11}$ .

Using the Pythagorean Theorem,

  $CD=\sqrt{1-a^2}$

The length of cathetus of cross-section triangle is equal to $2\sqrt{1-a^2}$ .

Finding the formulas for $A\left(y\right)$ on height $a\in \left[-1,1\right]$ perpendicular to the $y$ -axis,

  $A\left(a\right)=2-2a^2$

Finding the volume of the integral,

  $\begin{array}{c}V={\displaystyle \underset{-1}{\overset{1}{\int }}A\left(y\right)dy}\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}\left(2-2y^2\right)dy}\\ ={\left.\left(2y-\frac{2y^3}{3}\right)\right|}_{-1}^1\\ V=\frac{8}{3}\end{array}$

Therefore, the volume of the solid generated is $V=\frac{8}{3}\text{ cu}\text{. unit}$ .