← 2022 Paper 1

UPSC 2022 Maths Optional Paper 1 Q3b — Step-by-Step Solution

15 marks · Section A

Double integrals · Calculus · asked 10× in 13 yrs · Read the full method →

Question

Use double integration to calculate the area common to the circle $x^2+y^2=4$ and the parabola $y^2=3x$.

Technique

Intersection of curve and circle; bounded region described in terms of $y$ (horizontal slices); trig substitution for the circle integral; direct integration for the parabola term; double for symmetry.

Solution

Setup. The circle has radius 2 centred at origin. The parabola $y^2=3x$ opens to the right, with vertex at origin. The common region is inside the circle and to the right of (or “inside”) the parabola.

Actually, “common to” the circle (disc) and the parabola — both are curves, so we mean the region bounded by them. The region inside the circle ($x^2+y^2\le 4$) and to the right of the parabola ($y^2\le 3x$, i.e. the parabola opens to the right and the “inside” is the convex side).

Wait — clarify: $y^2=3x$ as a curve doesn’t bound a region by itself. The natural region “inside the parabola” is $y^2\le 3x$ (the convex region to the right of $y$-axis). The common region with disc $x^2+y^2\le 4$ is bounded.

Step 1 — Find intersection points

Solve $x^2+y^2=4$ and $y^2=3x$ simultaneously.

Substitute: $x^2+3x=4\Rightarrow x^2+3x-4=0\Rightarrow(x+4)(x-1)=0$. So $x=-4$ or $x=1$.

$x=-4$ requires $y^2=-12$, impossible. So $x=1$, then $y^2=3\Rightarrow y=\pm\sqrt 3$.

Intersections: $(1,\sqrt 3)$ and $(1,-\sqrt 3)$.

Step 2 — Describe the region

By symmetry about the $x$-axis, compute area for $y\ge 0$ and double.

For $0\le y\le\sqrt 3$: the region is bounded on the left by the parabola $x=y^2/3$ and on the right by the circle $x=\sqrt{4-y^2}$.

So $A_{\text{upper half}}=\int_0^{\sqrt 3}\!\left(\sqrt{4-y^2}-\dfrac{y^2}{3}\right)dy$.

Step 3 — Evaluate

Substitute $y=2\sin\theta$, $dy=2\cos\theta\,d\theta$. When $y=0,\theta=0$; when $y=\sqrt 3,\theta=\pi/3$ (since $\sin\pi/3=\sqrt 3/2$).

$\sqrt{4-y^2}=2\cos\theta$.

$I_1=\int_0^{\pi/3}2\cos\theta\cdot 2\cos\theta\,d\theta=4\int_0^{\pi/3}\cos^2\theta\,d\theta=4\int_0^{\pi/3}\dfrac{1+\cos 2\theta}{2}\,d\theta=2\left[\theta+\dfrac{\sin 2\theta}{2}\right]_0^{\pi/3}$ $=2\!\left[\dfrac{\pi}{3}+\dfrac{\sin(2\pi/3)}{2}-0\right]=2\!\left[\dfrac{\pi}{3}+\dfrac{\sqrt 3/2}{2}\right]=\dfrac{2\pi}{3}+\dfrac{\sqrt 3}{2}$.

$A_{\text{upper half}}=I_1-I_2=\dfrac{2\pi}{3}+\dfrac{\sqrt 3}{2}-\dfrac{\sqrt 3}{3}=\dfrac{2\pi}{3}+\dfrac{3\sqrt 3-2\sqrt 3}{6}=\dfrac{2\pi}{3}+\dfrac{\sqrt 3}{6}$.

Step 4 — Total area (double for symmetry)

Answer