GATE 2022 EE Q.51: solution and answer

GATE 2022 · EEQ.51 · 2 marks · MCQ

Let $R$ be a region in the first quadrant of the $x y$ plane enclosed by a closed curve $C$ considered in counter-clockwise direction. Which of the following expressions does not represent the area of the region $R$ ?

GATE 2022 EE question 51: The positive x and y axes meeting at the origin (first quadrant only). — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
$\iint_{R} d x d y$
(B)
$\oint_{C} x d y$
(C)
$\oint_{C} y d x$
(D)
$\frac{1}{2} \oint_{C} (x d y - y d x)$

Answer and solution

ANSWER

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Green's theorem, for a simple closed curve $C$ traversed counter-clockwise (positively oriented) around the region $R$ : $\oint_{C} (P d x + Q d y) = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y$ Any choice of $P, Q$ with $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ turns the line integral into the area of $R$ .
(A) $\iint_{R} d x d y$ is the area of $R$ by definition. It represents the area.
(B) Take $P = 0, Q = x$ . Then $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - 0 = 1$ , so $\oint_{C} x d y = \iint_{R} d x d y = Area of R$ It represents the area.
(C) Take $P = y, Q = 0$ . Then $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - 1 = - 1$ , so $\oint_{C} y d x = - \iint_{R} d x d y = - (Area of R)$ The area of a region is positive, so this is its negative. It does not represent the area.
(D) Take $P = - \frac{y}{2}, Q = \frac{x}{2}$ . Then $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{2} - (- \frac{1}{2}) = 1$ , so $\frac{1}{2} \oint_{C} (x d y - y d x) = \iint_{R} d x d y = Area of R$ It represents the area.
Only (C) fails: traversed counter-clockwise, $\oint_{C} y d x$ gives $- Area$ . The correct expression would be $- \oint_{C} y d x$ . Answer: (C).

KEY IDEA

By Green's theorem, $\oint_{C} (P d x + Q d y)$ taken counter-clockwise equals the area of $R$ exactly when $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ . That holds for $x d y$ and for $\frac{1}{2} (x d y - y d x)$ . For $y d x$ it is $- 1$ , which gives the negative of the area.

WATCH OUT FOR

Forgetting the minus sign in Green's theorem, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ , and so treating $\oint_{C} y d x$ as the area. Counter-clockwise, the area is $\oint_{C} x d y = - \oint_{C} y d x$ . A second mistake is to pick (D) because of its factor $\frac{1}{2}$ . The two terms each contribute half the area, so together they give the whole area.

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 23 Vector calculus questions, with notes

GATE 2022 EE Q.51: solution and answer

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