UPSC 2024 Maths Optional Paper 1 Q4b — Step-by-Step Solution

15 marks · Section A

Double integrals · Calculus · asked 11× in 14 yrs · Read the full method →

Question

Using double integration, find the area lying inside the cardioid $r=a(1+\cos\theta)$ and outside the circle $r=a$ .

Technique

Find intersection angles; set up a polar double integral with $r$ running from $a$ to the cardioid; exploit even symmetry.

Solution

$Cardioid r=a(1+\cos\theta) (red) and the circle r=a (blue). They meet where \cos\theta=0, i.e. at \theta=\pm\pi/2. The shaded required region is inside the cardioid and outside the circle; it occupies the angular range -\pi/2\le\theta\le\pi/2, where the cardioid lies outside the circle.$

Step 1 — Intersection angles.

$a(1+\cos\theta)=a\;\Rightarrow\;\cos\theta=0\;\Rightarrow\;\theta=\pm\pi/2$ .

For $\theta\in[-\pi/2,\pi/2]$ , $\cos\theta\ge 0$ so the cardioid $a(1+\cos\theta)\ge a$ : the required region lies in this angular range.

Step 2 — Area integral.

\text{Area}=\int_{-\pi/2}^{\pi/2}\!\!\int_{a}^{a(1+\cos\theta)}r\,dr\,d\theta.

Inner integral:

\int_{a}^{a(1+\cos\theta)}r\,dr=\frac{a^2}{2}\bigl[(1+\cos\theta)^2-1\bigr]=\frac{a^2}{2}(2\cos\theta+\cos^2\theta).

Both integrands are even, so by symmetry:

\text{Area}=a^2\int_0^{\pi/2}(2\cos\theta+\cos^2\theta)\,d\theta.

\int_0^{\pi/2}2\cos\theta\,d\theta=2,\qquad \int_0^{\pi/2}\cos^2\theta\,d\theta=\frac{\pi}{4}.

\text{Area}=a^2\Bigl(2+\frac{\pi}{4}\Bigr)=\frac{a^2(\pi+8)}{4}.

Answer

\boxed{\;\text{Area}=\frac{a^2(\pi+8)}{4}.\;}