UPSC Maths 2025 Paper 1 Q3c-i — Solution

10 marks · Section A

Question

Evaluate $\displaystyle\iint_R y\,dx\,dy$ , where $R$ is the region bounded by $y = x$ and $y = 4x - x^2$ .

Technique

Set up a double integral over the region between two curves: find the intersection points, use vertical strips with $y$ running from the lower to the upper curve, and integrate.

Solution

Step 1 — Find intersections.

Set $x = 4x - x^2$ : $x^2 - 3x = 0 \Rightarrow x(x-3)=0 \Rightarrow x=0,\ 3$ . So the curves meet at $(0,0)$ and $(3,3)$ .

Step 2 — Determine upper/lower curve on $[0,3]$ .

At $x=1$ : line gives $y=1$ , parabola gives $4-1=3$ . So the parabola $y=4x-x^2$ is above the line $y=x$ on $0<x<3$ .

Step 3 — Set up the iterated integral (vertical strips): $\iint_R y\,dx\,dy=\int_{x=0}^{3}\int_{y=x}^{\,4x-x^2} y\;dy\,dx.$

Step 4 — Inner integral. $\int_{x}^{4x-x^2} y\,dy=\left[\frac{y^2}{2}\right]_{x}^{4x-x^2}=\frac12\Big[(4x-x^2)^2-x^2\Big].$ Expand $(4x-x^2)^2=16x^2-8x^3+x^4$ , so $=\frac12\big(16x^2-8x^3+x^4-x^2\big)=\frac12\big(15x^2-8x^3+x^4\big).$

Step 5 — Outer integral. $\frac12\int_0^3\big(15x^2-8x^3+x^4\big)\,dx=\frac12\left[5x^3-2x^4+\frac{x^5}{5}\right]_0^3.$ At $x=3$ : $5(27)-2(81)+\dfrac{243}{5}=135-162+48.6=21.6=\dfrac{108}{5}$ . $=\frac12\cdot\frac{108}{5}=\frac{54}{5}.$

Answer

$\boxed{\;\iint_R y\,dx\,dy=\dfrac{54}{5}=10.8.\;}$