GATE 2018 IN Q.26: solution and answer

GATE 2018 · INQ.26 · 2 marks · MCQ

Given $F = (x^{2} - 2 y) i - 4 y z j + 4 x z^{2} k$ , the value of the line integral $\int_{c} F \cdot d l$ along the straight line $c$ from $(0, 0, 0)$ to $(1, 1, 1)$ is

(A)
$3/16$
(B)
$0$
(C)
$- 5/12$
(D)
$- 1$

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Answer and solution

ANSWER

Along the straight line from $(0, 0, 0)$ to $(1, 1, 1)$ , $\int_{c} F \cdot d l = - 1$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

Parametrise the straight line from $(0, 0, 0)$ to $(1, 1, 1)$ : $x = t, y = t, z = t$ with $0 \leq t \leq 1$ , so $d x = d y = d z = d t$ .
Write the line integral in component form: $\int_{c} F \cdot d l = \int_{c} [(x^{2} - 2 y) d x - 4 y z d y + 4 x z^{2} d z]$ .
Substitute the parametrisation into each component: $x^{2} - 2 y = t^{2} - 2 t$ , $- 4 y z = - 4 t^{2}$ , $4 x z^{2} = 4 t^{3}$ .
Since $d x = d y = d z = d t$ , add the three terms: $F \cdot d l = (t^{2} - 2 t - 4 t^{2} + 4 t^{3}) d t = (4 t^{3} - 3 t^{2} - 2 t) d t$ .
Integrate over $0 \leq t \leq 1$ : $\int_{0}^{1} (4 t^{3} - 3 t^{2} - 2 t) d t = [t^{4} - t^{3} - t^{2}]_{0}^{1} = 1 - 1 - 1 = - 1$
The field is not conservative ( $\nabla \times F \neq = 0$ ; for example the $k$ -component is $\frac{\partial F _{y}}{\partial x} - \frac{\partial F _{x}}{\partial y} = 0 - (- 2) = 2$ ), so the value depends on the path and holds for this straight line only. The answer is $- 1$ , option (D).

KEY IDEA

To evaluate a line integral along a given path, parametrise the path, here $x = y = z = t$ for $0 \leq t \leq 1$ . Write $F \cdot d l = F_{x} d x + F_{y} d y + F_{z} d z$ in terms of $t$ and integrate over the parameter range. Because $\nabla \times F \neq = 0$ , no potential function exists, so the integral has to be taken along this particular path.

WATCH OUT FOR

Dropping the minus signs in $- 2 y$ or $- 4 y z$ . Both give $t$ -terms of the same sign, and losing either one moves the result to a distractor. A second mistake is to assume $F$ is conservative and try to evaluate a potential at the endpoints: the curl is not zero, so the result depends on the path.

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GATE 2018 IN Q.26: solution and answer

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