Chapter 6.2, Problem 46E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 46E

The volume of the solid is $\frac{128}{3\sqrt{3}}$

Explanation of Solution

Given information: Given curve $y=\sqrt{16-x^2},-4\le x\le 4$

Calculation:

If we place the $x-$ axis along the diameter where the planes meet, then

The base of the solid is a semicircle with equation: $y=\sqrt{16-x^2},-4\le x\le 4$

A cross-section perpendicular to the $x-$ axis at a distance $x$ from the origin

Is a triangle $ABC$ .Base of triangle $ABC$ is $y=\sqrt{16-x^2}$ and whose height

  $\begin{array}{l}\left|BC\right|=y\tan {30}^0=\frac{\sqrt{16-x^2}}{\sqrt{3}}\\ \end{array}$

The cross-section is given by:

  $A\left(x\right)=\frac{1}{2}\sqrt{16-x^2}\cdot \frac{1}{\sqrt{3}}\sqrt{16-x^2}=\frac{16-x^2}{2\sqrt{3}}$

From previous volume is:

  $\begin{array}{l}V={\displaystyle \underset{-4}{\overset{4}{\int }}A\left(x\right)}dx\\ ={\displaystyle \underset{-4}{\overset{4}{\int }}\frac{16-x^2}{2\sqrt{3}}}dx\\ =\frac{1}{\sqrt{3}}{\displaystyle \underset{0}{\overset{4}{\int }}16-x^2}dx\\ =\frac{1}{\sqrt{3}}{\left[16x-\frac{x^3}{3}\right]}_0^4\\ =\frac{128}{3\sqrt{3}}\end{array}$

  

Thus, the volume of the solid is $\frac{128}{3\sqrt{3}}$