GATE 2023 EC Q.55: solution and answer

Official answer: 0Source: Official answer key, GATE 2023 EC (IIT Kanpur)

1. From the figure, $R$ is the square with vertices $(0,0)$, $(1,1)$, $(0,2)$ and $(-1,1)$. The lines $y=x$ and $y=-x+2$ meet at $(1,1)$, and $y=-x$ and $y=x+2$ meet at $(-1,1)$. The $y$-axis splits $R$ into two mirror-image halves.

2. Symmetry check: replacing $x$ by $-x$ turns $y=x$ into $y=-x$ and $y=-x+2$ into $y=x+2$, so $R$ is symmetric about the $y$-axis. The integrand $f(x,y)=xy$ is odd in $x$, since $f(-x,y)=-f(x,y)$. The two halves therefore cancel and the integral is $0$. Both halves are worked out below as a check.

3. Right half ($0\le x\le 1$, $x\le y\le 2-x$):
   $${\int }_0^{1}x{\left[\frac{y^2}{2}\right]}_x^{2-x}dx={\int }_0^{1}x\frac{(2-x{)}^2-x^2}{2}dx={\int }_0^{1}x\frac{4-4x}{2}dx={\int }_0^1(2x-2x^2)dx=1-\frac{2}{3}=\frac{1}{3}$$

4. Left half ($-1\le x\le 0$, $-x\le y\le x+2$):
   $${\int }_{-1}^{0}x\frac{(x+2{)}^2-x^2}{2}dx={\int }_{-1}^{0}x\frac{4x+4}{2}dx={\int }_{-1}^0(2x^2+2x)dx=\frac{2}{3}+(-1)=-\frac{1}{3}$$

5. Adding the halves: $\frac{1}{3}+\left(-\frac{1}{3}\right)=0$, so ${\iint }_{R}xydxdy=0$.