GATE 2022 EE Q.52: solution and answer

GATE 2022 · EEQ.52 · 2 marks · MCQ

Let $E (x, y, z) = 2 x^{2} \hat{i} + 5 y \hat{j} + 3 z \hat{k}$ . The value of $∭_{V} (\nabla \cdot E) d V$ , where $V$ is the volume enclosed by the unit cube defined by $0 \leq x \leq 1, 0 \leq y \leq 1$ , and $0 \leq z \leq 1$ , is

(A)
3
(B)
8
(C)
10
(D)
5

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Answer and solution

ANSWER

$∭_{V} (\nabla \cdot E) d V = 10$ , option (C).

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

The divergence of a vector field is $\nabla \cdot E = \frac{\partial E _{x}}{\partial x} + \frac{\partial E _{y}}{\partial y} + \frac{\partial E _{z}}{\partial z}$ . Here $E_{x} = 2 x^{2}$ , $E_{y} = 5 y$ and $E_{z} = 3 z$ .
$\frac{\partial}{\partial x} (2 x^{2}) = 4 x$ , $\frac{\partial}{\partial y} (5 y) = 5$ , $\frac{\partial}{\partial z} (3 z) = 3$ . So $\nabla \cdot E = 4 x + 8$ .
The integrand depends only on $x$ , and the cube has unit side, so the $y$ and $z$ integrals each contribute a factor of $1$ : $∭_{V} (4 x + 8) d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (4 x + 8) d x d y d z = \int_{0}^{1} (4 x + 8) d x .$
$\int_{0}^{1} (4 x + 8) d x = [2 x^{2} + 8 x]_{0}^{1} = 2 + 8 = 10.$
Check with the divergence theorem, which says the volume integral equals the net outward flux through the six faces. On the $x = 1$ face $E_{x} = 2 (1)^{2} = 2$ and on the $x = 0$ face $E_{x} = 0$ , so the net $x$ -flux is $2 - 0 = 2$ . The net $y$ -flux is $5 (1) - 5 (0) = 5$ and the net $z$ -flux is $3 (1) - 3 (0) = 3$ . Each face has area $1$ , so the total is $2 + 5 + 3 = 10$ , which matches. The answer is (C).

KEY IDEA

Work out the divergence $\nabla \cdot E = 4 x + 8$ and integrate it over the unit cube. You can check the result with the divergence theorem: the volume integral of the divergence equals the net outward flux through the cube's six faces, and here that is $2 + 5 + 3 = 10$ .

WATCH OUT FOR

Treating the divergence as the constant $2 + 5 + 3 = 10$ reaches the right number for the wrong reason. The $x$ -term is $4 x$ , not a constant, and it only averages to $2$ over $0 \leq x \leq 1$ . A second mistake is to differentiate $2 x^{2}$ and then set $x = 1$ , giving $4 + 5 + 3 = 12$ , which is not an option. A third is to add only the constant terms $5 + 3 = 8$ and drop the $x$ -term, which picks the distractor (B).

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GATE 2022 EE Q.52: solution and answer

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