Find the volume generated by revolving the area on the 1st and 4th quadrants by the ellipse y^2 / 9 + x^2 / 4 = 1 about the line x = 3.

$V = A \times 2\pi R$

$V = A \times 2\pi (3 - x_c)$
 

 

$V = \frac{1}{2}\pi ab \times 2\pi \left( 3 - \dfrac{4b}{3\pi} \right)$

$V = \frac{1}{2}\pi(3)(2) \times 2\pi \left( 3 - \dfrac{4(2)}{3\pi} \right)$

$V = 127.39 ~ \text{unit}^3$

$\dfrac{y^2}{9} + \dfrac{x^2}{4} = 1$

$\dfrac{x^2}{4} = 1 - \dfrac{y^2}{9}$

$x^2 = 4\left( 1 - \dfrac{y^2}{9} \right)$

$x = 2\sqrt{1 - \dfrac{y^2}{9}}$
 

 

$\displaystyle V = \int_{y_1}^{y_2} \pi (R^2 - r^2) \, dy$

$\displaystyle V = 2\pi \int_0^3 \left[ 3^2 - (3 - x)^2 \right] \, dy$

$\displaystyle V = 2\pi \int_0^3 \left[ 9 - \left( 3 - 2\sqrt{1 - \dfrac{y^2}{9}} \right)^2 \right] dy$

$V = 127.39 ~ \text{unit}^3$