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UPSC 2015 Maths Optional Paper 1 Q3d — Step-by-Step Solution

12 marks · Section A

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

Question

Evaluate ∬R(x−y)2cos⁡2(x+y) dx dy\displaystyle\iint_R (x-y)^2\cos^2(x+y)\,dx\,dy, where RR is the rhombus with successive vertices (π,0)(\pi,0), (2π,π)(2\pi,\pi), (π,2π)(\pi,2\pi), (0,π)(0,\pi).

Technique

Affine change of variables u=x+y, v=x−yu=x+y,\,v=x-y transforms a rhombus aligned with x±y=x\pm y= const into an axis-aligned rectangle; integrand factorises.

Solution

Strategy. The integrand depends on x−yx-y and x+yx+y. The rhombus is bounded by lines x+y=constx+y=\text{const} and x−y=constx-y=\text{const} (sides connect consecutive vertices). So substitute u=x+yu=x+y, v=x−yv=x-y.

Step 1 — Identify the sides of the rhombus

Successive vertices: A=(π,0)A=(\pi,0), B=(2π,π)B=(2\pi,\pi), C=(π,2π)C=(\pi,2\pi), D=(0,π)D=(0,\pi).

So in (u,v)=(x+y,x−y)(u,v)=(x+y,x-y) coordinates, RR is the rectangle [π,3π]×[−π,π][\pi,3\pi]\times[-\pi,\pi].

Step 2 — Change of variables

u=x+yu=x+y, v=x−yv=x-y. Then x=(u+v)/2x=(u+v)/2, y=(u−v)/2y=(u-v)/2.

Jacobian ∂(x,y)∂(u,v)=det⁡(1/21/21/2−1/2)=−1/2\dfrac{\partial(x,y)}{\partial(u,v)}=\det\begin{pmatrix}1/2 & 1/2\\ 1/2 & -1/2\end{pmatrix}=-1/2, so ∣J∣=1/2|J|=1/2.

dx dy=12 du dvdx\,dy=\tfrac{1}{2}\,du\,dv.

Step 3 — Rewrite integral

I=∬[π,3π]×[−π,π]v2cos⁡2u⋅12 du dv=12(∫π3πcos⁡2u du)(∫−ππv2 dv).I=\iint_{[\pi,3\pi]\times[-\pi,\pi]} v^2\cos^2 u\cdot\tfrac{1}{2}\,du\,dv=\tfrac{1}{2}\left(\int_\pi^{3\pi}\cos^2 u\,du\right)\left(\int_{-\pi}^{\pi}v^2\,dv\right).

Step 4 — Evaluate each integral

∫π3πcos⁡2u du=∫π3π1+cos⁡2u2 du=12[u+sin⁡2u2]π3π=12[(3π+0)−(π+0)]=12⋅2π=π.\int_\pi^{3\pi}\cos^2 u\,du=\int_\pi^{3\pi}\dfrac{1+\cos 2u}{2}\,du=\dfrac{1}{2}\left[u+\dfrac{\sin 2u}{2}\right]_\pi^{3\pi}=\dfrac{1}{2}\bigl[(3\pi+0)-(\pi+0)\bigr]=\dfrac{1}{2}\cdot 2\pi=\pi. ∫−ππv2 dv=[v33]−ππ=π33−−π33=2π33.\int_{-\pi}^{\pi}v^2\,dv=\left[\dfrac{v^3}{3}\right]_{-\pi}^{\pi}=\dfrac{\pi^3}{3}-\dfrac{-\pi^3}{3}=\dfrac{2\pi^3}{3}.

Step 5 — Combine

I=12⋅π⋅2π33=π43.I=\tfrac{1}{2}\cdot\pi\cdot\dfrac{2\pi^3}{3}=\dfrac{\pi^4}{3}.

Answer

  I=π43.  \boxed{\;I=\dfrac{\pi^4}{3}.\;}
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