Q. The volume of the solid obtained by revolving about $y$-axis the area enclosed between the ellipse $x^{2} + 9y^{2} = 9$ and the straight line $x + 3y = 3,\$in the first quadrant is
A) $3\pi$
B) 4$\ \pi$
C) 6$\ \pi$
D) 9$\ \pi$
Explanation
(a)
$Volume = \int_{0}^{1}\pi\ x^{2}\ dy - \int_{0}^{1}\pi\ x^{2}\ dy$
$= \pi\int_{0}^{1}{9\left( 1 - x^{2} \right)dy - \pi\int_{0}^{1}{9(1 - {y)}^{2}\ dy}}$
$= 9\pi\left\lbrack \left( y - \frac{y^{3}}{3} \right) + \frac{(1 - y)^{3}}{3} \right\rbrack_{0}^{1} = 9\pi\left\lbrack \left( 1 - \frac{1}{3} \right) + \left( 0 - \frac{1}{3} \right) \right\rbrack = 3\pi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.