UPSC 2018 Maths Optional Paper 1 Q2c — Step-by-Step Solution

13 marks · Section A

Areas, surface areas, volumes via integration · Calculus · asked 4× in 13 yrs · Read the full method →

Question

The ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ revolves about the $x$ -axis. Find the volume of the solid of revolution.

Technique

Disc (washer with inner radius $0$ ) method about the $x$ -axis; $V=\pi\int y^2\,dx$ .

Solution

Step 1 — Disc method setup

Revolving the region bounded by the ellipse about the $x$ -axis produces a prolate/oblate spheroid. At station $x\in[-a,a]$ the cross-section perpendicular to the axis is a disc of radius $y$ , where

y^2=b^2\!\left(1-\frac{x^2}{a^2}\right).

The volume is

V=\int_{-a}^{a}\pi y^2\,dx=\pi b^2\int_{-a}^{a}\!\left(1-\frac{x^2}{a^2}\right)dx.

Step 2 — Evaluate the integral

By symmetry, $V=2\pi b^2\displaystyle\int_0^a\!\left(1-\frac{x^2}{a^2}\right)dx$ . Now

\int_0^a\!\left(1-\frac{x^2}{a^2}\right)dx=\Big[x-\frac{x^3}{3a^2}\Big]_0^a=a-\frac{a}{3}=\frac{2a}{3}.

Hence

V=2\pi b^2\cdot\frac{2a}{3}=\frac{4}{3}\pi a b^2.

Answer

\boxed{\;V=\frac{4}{3}\pi a b^{2}.\;}