Chapter A5.1, Problem 42E

To determine

To find: the volume of the solid.

Answer to Problem 42E

The volume of the solid is $V=24\pi$ .

Explanation of Solution

Given:

The line $x=4,$ and the hyperbola $9x^2-4y^2=36$ .

Calculation:

Equation of hyperbola: $9x^2-4y^2=36$ .

This implies that,

  $\begin{array}{l}{y}^2=\frac{9}{4}{x}^2-9\\ y=\pm \sqrt{\frac{9}{4}{x}^2-9}\end{array}$

Since $y$ cannot be negative, use the positive root on the interval $2\le x\le 4$ .

  $\begin{array}{c}V={\displaystyle {\int }_2^4\pi }{\left(\sqrt{\frac{9}{4}{x}^2-9}\right)}^2\cdot dx\\ ={\displaystyle {\int }_2^4\pi }{\left(\sqrt{\frac{9}{4}{x}^2-9}\right)}^{2}dx\\ =\pi {\left[\frac{3}{4}{x}^3-9x\right]}_2^4\\ =24\pi \end{array}$

Therefore, $V=24\pi$ .

Conclusion:

Therefore, the volume of the solid is $V=24\pi$ .