GATE 2024 IN Q.46: solution and answer

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Write the integral as ${\oint }_C(Pdx+Qdy)$ with $P=-e^y$ and $Q=e^x$. $C$ is the boundary of the rectangle, a simple closed curve traversed counterclockwise, and $P$, $Q$ have continuous partial derivatives everywhere, so Green's theorem applies: $${\oint }_C(Pdx+Qdy)={\iint }_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$$

2. Partial derivatives: $\frac{\partial Q}{\partial x}=e^x$ and $\frac{\partial P}{\partial y}=-e^y$, so the integrand is $e^x-(-e^y)=e^x+e^y$.

3. The region is the rectangle $R:0\le x\le 2,0\le y\le 1$. Split the double integral into two parts: $${\iint }_R(e^x+e^y)dA={\int }_0^1{\int }_0^2{e}^{x}dxdy+{\int }_0^2{\int }_0^1{e}^{y}dydx$$

4. First term: $\left({\int }_0^2{e}^{x}dx\right)(1)=e^2-1$.

5. Second term: $(2)\left({\int }_0^1{e}^{y}dy\right)=2(e-1)=2e-2$.

6. Add them: $(e^2-1)+(2e-2)=e^2+2e-3$. This is option (A).

7. Check with the direct line integral. On the bottom edge ($y=0$, $x$: $0\to 2$, $dy=0$) the integral is ${\int }_0^2-1dx=-2$. On the right edge ($x=2$, $y$: $0\to 1$, $dx=0$) it is ${\int }_0^1{e}^{2}dy=e^2$. On the top edge ($y=1$, $x$: $2\to 0$, $dy=0$) it is ${\int }_2^0-edx=2e$. On the left edge ($x=0$, $y$: $1\to 0$, $dx=0$) it is ${\int }_1^{0}1dy=-1$. The total is $-2+e^2+2e-1=e^2+2e-3$, which agrees.