The part of circle $x^{2} + y^{2} = 9$ in between $y = 0$ and $y = 2$ is revolved about $y$-axis. The volume of generating solid will be

Question

Dr. Abubacker · Accepted Answer

(a)
Volume of generated solid
$= \pi\int_{0}^{2}{x^{2}\ \ dy = \pi\int_{0}^{2}{\left( 9 - y^{2} \right)dy = \pi\left\lbrack 9y - \frac{1}{3}y^{3} \right\rbrack_{0}^{2}}}$
$\pi = \left\lbrack 18 - \frac{8}{3} \right\rbrack = \frac{46}{3}\pi\ cu\ units$

Q. The part of circle $x^{2} + y^{2} = 9$ in between $y = 0$ and $y = 2$ is revolved about $y$-axis. The volume of generating solid will be

Explanation

Reviewed by Dr. Abubacker

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