Chapter 8.3, Problem 13E

To determine

To find : the volume of the solid.

Answer to Problem 13E

The volume of the solid is $36\pi$ .

Explanation of Solution

Given information : The region bounded by the lines and curves about the x -axis $y=\sqrt{9-x^2},y=0$ .

Formula used: Area $=\pi r^2$

Calculation :

The curve $y=\sqrt{9-x^2}$

Area $=\pi r^2$

Area $=\pi {\left(\sqrt{9-x^2}\right)}^2$

The volume of the solid:

  $\begin{array}{l}V={\displaystyle \underset{-3}{\overset{3}{\int }}\pi r^2}dx\\ V={\displaystyle \underset{-3}{\overset{3}{\int }}\pi {\left(\sqrt{9-x^2}\right)}^2}dx\\ V=\pi {\displaystyle \underset{-3}{\overset{3}{\int }}{\left(\sqrt{9-x^2}\right)}^2}dx\\ V=\pi {\displaystyle \underset{-3}{\overset{3}{\int }}\left(9-x^2\right)}dx\\ V=\pi {\left[9x-\frac{x^3}{3}\right]}_{-3}^3\\ V=\pi \left[9\left(3\right)-\frac{{\left(3\right)}^3}{3}\right]-\left[9\left(-3\right)-\frac{{\left(-3\right)}^3}{3}\right]\\ V=\pi \left(27-9\right)-\left(-27-\left(-9\right)\right)\\ V=36\pi \end{array}$