Question

The curl of vector field \(\vec{f} (x,y,z) = x^2\hat{i} + 2z \hat{j} – y \hat{k} \) is  _________

(a) \(-3\hat{i} \)

(b) \(-3\hat{j} \)

(c) \(-3\hat{k} \)

(d) 0

I had been asked this question during an online exam.

This intriguing question comes from Divergence and Curl of a Vector Field topic in division Vector Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (a) \(-3\hat{i} \)

Explanation: Curl,

\(\vec{f} = ∇ ⤫ \vec{f} =  \begin{vmatrix}

i & j & k\\

\frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\

x^2 & 2z & -y\\

\end{vmatrix} \)

\( = \bigg\{\frac{∂}{∂y} (-y)-\frac{∂}{∂z}(2z)) \hat{i} + (\frac{∂}{∂z} (x^2 )-\frac{∂}{∂x}(-y)) \hat{j} + (\frac{∂}{∂x} (2z)-\frac{∂}{∂y} (x^2 ) \hat{k}\bigg\} \)

\( = (-1-2) \hat{i} + (0-0) \hat{j} + (0-0) \hat{k} = -3\hat{i}. \)