GATE 2022 EC Q.11: solution and answer

GATE 2022 · ECQ.11 · 1 mark · MCQ

Consider the two-dimensional vector field $F (x, y) = x  + y $ , where $$ and $$ denote the unit vectors along the $x$ -axis and the $y$ -axis, respectively. A contour $C$ in the $x$ - $y$ plane, as shown in the figure, is composed of two horizontal lines connected at the two ends by two semicircular arcs of unit radius. The contour is traversed in the counter-clockwise sense. The value of the closed path integral

\oint_{C} F (x, y) \cdot (d x  + d y )

is ________.

GATE 2022 EC question 11: x-y axes with the origin marked (0,0). — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$0$
(B)
$1$
(C)
$8 + 2 π$
(D)
$- 1$

Answer and solution

ANSWER

The closed path integral is $0$ , which is option (A). The field $F = x  + y $ is the gradient of $ϕ = \frac{x ^{2} + y ^{2}}{2}$ , so its integral around any closed curve is zero.

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Write the field as $F = P  + Q $ with $P = x$ , $Q = y$ . With $d r = d x  + d y $ , the integrand is $F \cdot d r = x d x + y d y$ .
Test whether the field is conservative (2-D curl): $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - 0 = 0$ everywhere, and $P$ , $Q$ are smooth on the whole plane, which is simply connected. So $F$ is a gradient field.
Find the potential: $x d x + y d y = d ϕ$ with $ϕ (x, y) = \frac{x ^{2} + y ^{2}}{2}$ . Check: $\nabla ϕ = x  + y  = F$ .
Fundamental theorem for line integrals: for a gradient field, $\int \nabla ϕ \cdot d r = ϕ (end) - ϕ (start)$ . A closed path ends where it starts, so $\oint_{C} \nabla ϕ \cdot d r = 0$ .
Cross-check with Green's theorem ( $C$ is a simple, piecewise-smooth closed curve traversed counter-clockwise): $\oint_{C} (P d x + Q d y) = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A = \iint_{R} 0 d A = 0.$ So the shape of the contour, the arc radii and the direction of traversal make no difference.
Direct check, piece by piece, using $ϕ$ at the four junctions: $ϕ (0, 0) = 0$ , $ϕ (4, 0) = 8$ , $ϕ (4, 2) = 10$ , $ϕ (0, 2) = 2$ . Bottom line: $8 - 0 = 8$ . Right arc: $10 - 8 = 2$ . Top line: $2 - 10 = - 8$ . Left arc: $0 - 2 = - 2$ . Total: $8 + 2 - 8 - 2 = 0$ .
Hence $\oint_{C} F \cdot (d x  + d y ) = 0$ , which is option (A).

KEY IDEA

$F = x  + y  = \nabla (\frac{x ^{2} + y ^{2}}{2})$ is a conservative (gradient) field: its 2-D curl $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is zero everywhere. The line integral of a gradient field depends only on where the path starts and ends, so around any closed contour it is $0$ . The figure's geometry is a distraction.

WATCH OUT FOR

Picking (C), $8 + 2 π$ . That is the length of $C$ (two lines of length $4$ plus two semicircles of length $π$ ), which is $\oint_{C} ∣ d r ∣$ , not the work integral $\oint_{C} F \cdot d r$ . The other trap is working through the four pieces and getting a direction wrong on the top line or an arc, so the pieces fail to cancel. Check the curl first: once it is zero on the whole plane, the closed-path integral is $0$ with no parametrisation needed.

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

GATE 2022 EC Q.11: solution and answer

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