GATE EE 2022 Set 1 — Question 51

The area of a region $R$ enclosed by a simple closed curve $C$ (traversed counter-clockwise) can be expressed using various integral forms. The most fundamental definition of the area of a region $R$ is given by the double integral:
$\text{Area} = \iint_R dA = \iint_R dxdy$
Thus, option (A) correctly represents the area. Green's Theorem states that for a region $R$ enclosed by a positively oriented (counter-clockwise) simple closed curve $C$, if $P(x,y)$ and $Q(x,y)$ have continuous partial derivatives on an open region containing $R$, then:
$\oint_C (Pdx + Qdy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$
To represent the area, we need the integrand of the double integral to be 1, i.e., $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$. Let's check the other options using Green's Theorem:

• Option (B): $\oint_C xdy$

Here, $P=0$ and $Q=x$. So, $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = 0$.
$\oint_C xdy = \iint_R (1 - 0) dA = \iint_R dA = \text{Area}$.
Thus, option (B) correctly represents the area.

• Option (C): $\oint_C ydx$

Here, $P=y$ and $Q=0$. So, $\frac{\partial Q}{\partial x} = 0$ and $\frac{\partial P}{\partial y} = 1$.
$\oint_C ydx = \iint_R (0 - 1) dA = -\iint_R dA = -\text{Area}$.
Thus, option (C) represents the negative of the area, not the area itself.

• Option (D): $\frac{1}{2} \oint_C (xdy - ydx)$

This can be written as $\oint_C \left( -\frac{y}{2} dx + \frac{x}{2} dy \right)$.
Here, $P = -\frac{y}{2}$ and $Q = \frac{x}{2}$. So, $\frac{\partial Q}{\partial x} = \frac{1}{2}$ and $\frac{\partial P}{\partial y} = -\frac{1}{2}$.
$\frac{1}{2} \oint_C (xdy - ydx) = \iint_R \left( \frac{1}{2} - \left(-\frac{1}{2}\right) \right) dA = \iint_R (\frac{1}{2} + \frac{1}{2}) dA = \iint_R 1 dA = \text{Area}$.
Thus, option (D) correctly represents the area.The question asks which expression does not represent the area of the region $R$. Based on the analysis, option (C) does not represent the area.The final answer is $\boxed{C}$