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UPSC 2017 Maths Optional Paper 1 Q4d — Step-by-Step Solution

10 marks · Section A

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

Question

Prove that $\dfrac{\pi}{3}\le\displaystyle\iint_D\frac{dx\,dy}{\sqrt{x^2+(y-2)^2}}\le\pi$ where $D$ is the unit disc.

Technique

Recognize the integrand as $1/\text{dist}(\cdot,Q)$; bound the distance by $[\,d(O,Q)-r,\ d(O,Q)+r\,]=[1,3]$; reciprocate and multiply by $\operatorname{area}(D)=\pi$.

Solution

Step 1 — Geometric meaning of the integrand

The integrand is

So $g$ is the reciprocal distance from the variable point $(x,y)\in D$ to the fixed point $Q=(0,2)$, which lies outside the unit disc $D=\{x^2+y^2\le1\}$.

Step 2 — Extreme distances from $Q$ to points of $D$

$Q=(0,2)$ is at distance $2$ from the centre $O$ of the disc, and $D$ has radius $1$. For any point $P\in D$,

The minimum $1$ is attained at $(0,1)$ (nearest point of $D$ to $Q$) and the maximum $3$ at $(0,-1)$.

Step 3 — Bound the integrand pointwise

Taking reciprocals (distance is positive on $D$ since $Q\notin D$):

Step 4 — Integrate the inequalities over $D$

The area of the unit disc is $\iint_D dx\,dy=\pi$. Monotonicity of the integral gives

that is

Answer