← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

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

Question

Integrate the function f(x,y)=xy(x2+y2)f(x,y)=xy(x^2+y^2) over the domain R:{−3≤x2−y2≤3, 1≤xy≤4}R:\{-3\le x^2-y^2\le 3,\ 1\le xy\le 4\}.

Technique

Map the region to a rectangle with u=x2−y2, v=xyu=x^2-y^2,\ v=xy; the Jacobian 2(x2+y2)2(x^2+y^2) cancels the (x2+y2)(x^2+y^2) in ff, leaving an elementary rectangle integral.

Solution

Step 1 — Choose the natural substitution

The region is described by the level sets of x2−y2x^2-y^2 and xyxy, so set

u=x2−y2,v=xy.u=x^2-y^2,\qquad v=xy.

In the (u,v)(u,v) plane the region is the rectangle

−3≤u≤3,1≤v≤4.-3\le u\le 3,\qquad 1\le v\le 4.

Step 2 — Jacobian of (u,v)(u,v) with respect to (x,y)(x,y)

∂(u,v)∂(x,y)=∣uxuyvxvy∣=∣2x−2yyx∣=2x2+2y2=2(x2+y2).\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}u_x & u_y\\ v_x & v_y\end{vmatrix}=\begin{vmatrix}2x & -2y\\ y & x\end{vmatrix}=2x^2+2y^2=2(x^2+y^2).

Hence

du dv=2(x2+y2) dx dy⟹dx dy=du dv2(x2+y2).du\,dv=2(x^2+y^2)\,dx\,dy\quad\Longrightarrow\quad dx\,dy=\frac{du\,dv}{2(x^2+y^2)}.

(Since 1≤v=xy1\le v=xy, we stay in a region where xy>0xy>0 and x2+y2>0x^2+y^2>0, so the Jacobian never vanishes and the map is locally invertible.)

Step 3 — Transform the integrand

f dx dy=xy(x2+y2)⋅du dv2(x2+y2)=xy2 du dv=v2 du dv.f\,dx\,dy=xy(x^2+y^2)\cdot\frac{du\,dv}{2(x^2+y^2)}=\frac{xy}{2}\,du\,dv=\frac{v}{2}\,du\,dv.

The factor (x2+y2)(x^2+y^2) cancels cleanly — this is precisely why this substitution is the right one.

Step 4 — Integrate over the rectangle

∬Rf dx dy=∫u=−33∫v=14v2 dv du=∫−33[v24]14du=∫−3316−14 du=154⋅6.\iint_R f\,dx\,dy=\int_{u=-3}^{3}\int_{v=1}^{4}\frac{v}{2}\,dv\,du=\int_{-3}^{3}\left[\frac{v^2}{4}\right]_{1}^{4}du=\int_{-3}^{3}\frac{16-1}{4}\,du=\frac{15}{4}\cdot 6.

Answer

  ∬Rxy(x2+y2) dx dy=452.  \boxed{\;\iint_R xy(x^2+y^2)\,dx\,dy=\frac{45}{2}.\;}
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