GATE 2023 EE Q.62: solution and answer

GATE 2023 · EEQ.62 · 2 marks · NAT

The closed curve shown in the figure is described by

$r = 1 + cos θ$ , where $r = x^{2} + y^{2}$ ; $x = r cos θ$ , $y = r sin θ$

The magnitude of the line integral of the vector field $F = - y \hat{i} + x \hat{j}$ around the closed curve is ______ (Round off to 2 decimal places).

GATE 2023 EE question 62: Plot of a cardioid on x-y axes. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The magnitude of the line integral is $3 π \approx 9.42$ .

Official answer: 9 to 10Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Write the line integral in component form. With $d r = d x \hat{i} + d y \hat{j}$ : $\oint_{C} F \cdot d r = \oint_{C} (- y d x + x d y) .$
The cardioid is a simple closed curve. Its only corner is the cusp at the origin, so it is piecewise smooth. $F$ is smooth everywhere, so Green's theorem applies. With $P = - y$ and $Q = x$ : $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (- 1) = 2,$ so $\oint_{C} (x d y - y d x) = \iint_{R} 2 d A = 2 \times (area enclosed by C) .$
Find the area in polar coordinates: $A = \frac{1}{2} \int_{0}^{2 π} r^{2} d θ = \frac{1}{2} \int_{0}^{2 π} (1 + cos θ)^{2} d θ .$
Expand the integrand: $(1 + cos θ)^{2} = 1 + 2 cos θ + cos^{2} θ$ . Over one full period, $\int_{0}^{2 π} 1 d θ = 2 π$ , $\int_{0}^{2 π} cos θ d θ = 0$ and $\int_{0}^{2 π} cos^{2} θ d θ = π$ . So $\int_{0}^{2 π} (1 + cos θ)^{2} d θ = 3 π$ .
So the area is $A = \frac{3 π}{2}$ .
The line integral is $2 A = 3 π \approx 9.4248$ . You can check this directly: with $x = r cos θ$ and $y = r sin θ$ , $x d y - y d x = r^{2} d θ$ , so the integral is $\int_{0}^{2 π} (1 + cos θ)^{2} d θ = 3 π$ .
Going around the curve counter-clockwise gives $+ 3 π$ and going clockwise gives $- 3 π$ . The question asks for the magnitude, which is $3 π \approx 9.42$ .

KEY IDEA

For $F = - y \hat{i} + x \hat{j}$ , Green's theorem gives $\oint_{C} (x d y - y d x) = 2 \times (enclosed area)$ . The problem therefore reduces to the area of the cardioid $r = 1 + cos θ$ , which is $\frac{1}{2} \int_{0}^{2 π} r^{2} d θ = \frac{3 π}{2}$ . The same result follows directly from $x d y - y d x = r^{2} d θ$ .

WATCH OUT FOR

The usual slip is reporting the area $3 π /2 \approx 4.71$ instead of twice the area. The curl term is $1 - (- 1) = 2$ , not 1. The second most common error is forgetting the factor $\frac{1}{2}$ in the polar area formula, which gives $6 π$ . Other errors: dropping the $cos^{2} θ$ term, which gives $2 π$ , or integrating only over $0$ to $π$ , which covers half the curve.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 23 Vector calculus questions, with notes

GATE 2023 EE Q.62: solution and answer

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