Question

Volume of solid obtained by revolving the area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about major and minor axes are in the ratio

• A. $b^2:a^2$
• B. $a^2:b^2$
• C. $a:b$
• D. $b:a$

Answer 1

Answer: (D)

$b:a$

Case (i): When ellipse is rotated about major axis:

Take a small disc at a length of $x$ from the centre of thickness $dx$. Then the volume of solid obtained by rotation will be ${\int }_{-a}^a\left(\text{Area}\right)dx$

Area of disc $=\pi r^2$

$r$ can be calculated from the equation of ellipse as

$\frac{x^2}{a^2}+\frac{r^2}{b^2}=1$

$\Rightarrow r^2=b^2(1-\frac{x^2}{a^2})=\frac{b^2}{a^2}(a^2-x^2)$

$\therefore {\text{Volume}}_{\text{major axis}}={\int }_{-a}^a\pi \frac{b^2}{a^2}(a^2-x^2)dx={[\pi b^{2}x-\frac{\pi b^2{x}^3}{3a^2}]}_{-a}^a=\frac{4}{3}\pi a{b}^2$

Case (ii): When ellipse is rotated about minor axis:

Following similar procedure as case (i),

$r^2=\frac{a^2}{b^2}(b^2-y^2)$

In this case, the area will be integrated w.r.t $dy$ as it is rotated about the $Y$-axis.

$\therefore {\text{Volume}}_{\text{minor axis}}={\int }_{-b}^b\pi \frac{a^2}{b^2}(b^2-y^2)dy={[\pi a^{2}y-\frac{\pi a^2{y}^3}{3b^2}]}_{-b}^b=\frac{4}{3}\pi a^{2}b$

$\therefore \frac{\text{Volume about major axis}}{\text{Volume about minor axis}}=\frac{\frac{4}{3}\pi a{b}^2}{\frac{4}{3}\pi a^{2}b}=\frac{b}{a}$