GATE 2026 EC Q.61: solution and answer

Official answer: 0.60 to 0.70Source: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Read the region. The figure puts the left and right corners of $R$ on the $x$-axis at $(0,0)$ and $(2,0)$, so that diagonal has length $2$ and midpoint $(1,0)$. In a square the diagonals are equal and are perpendicular bisectors of each other, so the dotted vertical diagonal through $x=1$ also has length $2$. The other two corners are therefore $(1,1)$ and $(1,-1)$. The side is $a=\sqrt{2}$, the area is $a^2=2$, and $R$ is the set $|x-1|+|y|\le 1$.

2. Move the origin to the centre with $u=x-1$ (a translation, so the Jacobian is $1$). Then $x^2+y^2-1=(1+u{)}^2+y^2-1=u^2+2u+y^2$.

3. The region $|u|+|y|\le 1$ is symmetric under $u\to -u$ and $2u$ is odd in $u$, so ${\iint }_{R}2udA=0$.

4. That leaves $I={\iint }_R(u^2+y^2)dA$ over the diamond $|u|+|y|\le 1$. For a fixed $y$ with $|y|\le 1$, $u$ runs from $-(1-|y|)$ to $1-|y|$, a width of $2(1-|y|)$.

5. $${\iint }_R{y}^{2}dA={\int }_{-1}^1{y}^2\cdot 2(1-|y|)dy=4{\int }_0^1(y^2-y^3)dy=4\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{3}$$

6. The diamond is unchanged when $u$ and $y$ are swapped, so ${\iint }_R{u}^{2}dA=\frac{1}{3}$ as well.

7. $$I=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\approx 0.67$$

8. Check: $\iint (u^2+y^2)dA$ is the polar moment of area of a square about its centre, which is $\frac{a^4}{6}$ and does not change when the square is rotated. Here $\frac{(\sqrt{2}{)}^4}{6}=\frac{4}{6}=\frac{2}{3}$, which agrees.