Chapter A5.1, Problem 41E

(a)

To determine

To find: the volumes of the solid about the x-axis.

Answer to Problem 41E

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

Explanation of Solution

Given:

Equation of ellipse: $9x^2+4y^2=36$

Calculation:

Equation of ellipse: $9x^2+4y^2=36$

This implies that,

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

Since $y$ cannot be negative, use the positive root.

  $\begin{array}{c}V=2{\displaystyle {\int }_0^2\pi }{\left(\sqrt{9-\frac{9}{4}{x}^2}\right)}^2\cdot dx\\ =2{\displaystyle {\int }_0^2\pi }{\left(\sqrt{9-\frac{9}{4}{x}^2}\right)}^{2}dx\\ =2\pi {\left[9x-\frac{3}{4}{x}^3\right]}_0^2\\ =24\pi \end{array}$

Therefore, the volume $V=24\pi$ .

Conclusion:

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

(b)

To determine

To find: the volumes of the solid about the y-axis.

Answer to Problem 41E

The volume of the solid is $V=16\pi$

Explanation of Solution

Given:

Equation of ellipse: $9x^2+4y^2=36$

Calculation:

Equation of ellipse: $9x^2+4y^2=36$

This implies that,

  $x^2=4-\frac{4}{9}{y}^2$

  $x=\pm \sqrt{4-\frac{4}{9}{y}^2}$

Since y cannot be negative, use the positive root

  $\begin{array}{c}V=2{\displaystyle {\int }_0^3\pi }{\left(\sqrt{4-\frac{4}{9}{y}^2}\right)}^{2}dy\\ =2{\displaystyle {\int }_0^2\pi }{\left(\sqrt{4-\frac{4}{9}{y}^2}\right)}^{2}dy\\ =2\pi {\left[4y-\frac{4}{27}{y}^3\right]}_0^3\\ =16\pi \end{array}$

Therefore, the volume $V=16\pi$

Conclusion:

Therefore, the volume of the solid is $V=16\pi$ .