← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q4b — Step-by-Step Solution

15 marks · Section A

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

Question

Show that the entire area of the astroid $x^{2/3}+y^{2/3}=a^{2/3}$ is $\dfrac{3}{8}\pi a^2$.

Technique

Parametric area formula $\oint x\,dy$; Beta function reduction.

Solution

Parametric form. The astroid has the standard parametrization:

Verify: $x^{2/3}=a^{2/3}\cos^2\theta$, $y^{2/3}=a^{2/3}\sin^2\theta$, sum = $a^{2/3}(\cos^2+\sin^2)=a^{2/3}$ ✓.

Step 1 — Area formula (parametric)

$A=\oint x\,dy=\int_0^{2\pi}x(\theta)\cdot\dfrac{dy}{d\theta}\,d\theta$.

$\dfrac{dy}{d\theta}=3a\sin^2\theta\cos\theta$.

$A=\int_0^{2\pi}a\cos^3\theta\cdot 3a\sin^2\theta\cos\theta\,d\theta=3a^2\int_0^{2\pi}\cos^4\theta\sin^2\theta\,d\theta$.

Step 2 — Use symmetry: $4\times$ first quadrant

By symmetry: $\int_0^{2\pi}\cos^4\theta\sin^2\theta\,d\theta=4\int_0^{\pi/2}\cos^4\theta\sin^2\theta\,d\theta$.

(All four “quadrants” contribute equally.)

Step 3 — Beta function

$\int_0^{\pi/2}\cos^m\theta\sin^n\theta\,d\theta=\dfrac{1}{2}B\!\left(\dfrac{m+1}{2},\dfrac{n+1}{2}\right)=\dfrac{1}{2}\cdot\dfrac{\Gamma((m+1)/2)\Gamma((n+1)/2)}{\Gamma((m+n+2)/2)}$.

With $m=4,n=2$: $=\dfrac{1}{2}\cdot\dfrac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)}=\dfrac{1}{2}\cdot\dfrac{(3/2)(1/2)\sqrt\pi\cdot(1/2)\sqrt\pi}{6}=\dfrac{1}{2}\cdot\dfrac{(3/8)\pi}{6}=\dfrac{3\pi}{96}=\dfrac{\pi}{32}$.

Step 4 — Compute area

$A=3a^2\cdot 4\cdot\dfrac{\pi}{32}=\dfrac{12a^2\pi}{32}=\dfrac{3\pi a^2}{8}$.

Answer