← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q4b — Step-by-Step Solution

12 marks · Section A

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

Question

Evaluate the integral ∫0a∫x/axx dy dxx2+y2\displaystyle\int_0^a\int_{x/a}^x \frac{x\,dy\,dx}{x^2+y^2}.

(Here a>0a>0; for the inner limits to be ordered we take a≥1a\ge1.)

Technique

Inner yy-integral via ∫dy/(x2+y2)=1xarctan⁡(y/x)\int dy/(x^2+y^2)=\frac1x\arctan(y/x); the xx in the numerator cancels the 1/x1/x, leaving an xx-independent constant that integrates trivially.

Solution

Step 1 — Inner integral over yy

Treat xx as constant. Since ∫dyx2+y2=1xarctan⁡yx\displaystyle\int\frac{dy}{x^2+y^2}=\frac1x\arctan\frac yx,

∫x/axx dyx2+y2=x⋅1x[arctan⁡yx]y=x/ay=x=arctan⁡xx−arctan⁡x/ax=arctan⁡1−arctan⁡1a.\int_{x/a}^{x}\frac{x\,dy}{x^2+y^2}=x\cdot\frac1x\Big[\arctan\frac yx\Big]_{y=x/a}^{y=x}=\arctan\frac{x}{x}-\arctan\frac{x/a}{x}=\arctan1-\arctan\frac1a.

So the inner integral is the constant π4−arctan⁡1a\dfrac\pi4-\arctan\dfrac1a (independent of xx).

Step 2 — Outer integral over xx

∫0a ⁣(π4−arctan⁡1a)dx=(π4−arctan⁡1a) ⁣∫0adx=a ⁣(π4−arctan⁡1a).\int_0^a\!\left(\frac\pi4-\arctan\frac1a\right)dx=\left(\frac\pi4-\arctan\frac1a\right)\!\int_0^a dx=a\!\left(\frac\pi4-\arctan\frac1a\right).

Answer

  ∫0a∫x/axx dy dxx2+y2=a(π4−arctan⁡1a)=πa4−aarctan⁡1a.  \boxed{\;\int_0^a\int_{x/a}^x \frac{x\,dy\,dx}{x^2+y^2}=a\left(\frac{\pi}{4}-\arctan\frac{1}{a}\right)=\frac{\pi a}{4}-a\arctan\frac1a.\;}
We post more of this — worked solutions, CSAT trap breakdowns, guide chapters — a few times a week on Telegram. Free, no sign-in. Join

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.