Question

Chose the curl of \(\vec{f} (x ,y ,z) = x^2 \hat{i} + xyz \hat{j} – z \hat{k} \) at the point (2, 1, -2).

(a) \(2\hat{i} + 2\hat{k} \)

(b) \(-2\hat{i} – 2\hat{j} \)

(c) \(4\hat{i} – 4\hat{j} + 2\hat{k} \)

(d) \(-2\hat{i} – 2\hat{k} \)

I have been asked this question in exam.

This interesting question is from Divergence and Curl of a Vector Field in portion Vector Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (d) \(-2\hat{i} – 2\hat{k} \)

The explanation: \(\vec{f} (x ,y ,z) = x^2 \hat{i} + xyz \hat{j} – z \hat{k} \)

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

i & j & k\\

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

x^2 & xyz & -z\\

\end{vmatrix} \)

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

\( = (0-xy) \hat{i} + (0-0) \hat{j} + (yz-0) \hat{k} \)

\( = – xy \hat{i} + yz \hat{k} \)

\(∇ ⤫ \vec{f} |_{(1, -1, 2)} = -2(1) \hat{i} + (1) (-2) \hat{k} = -2\hat{i} – 2\hat{k}. \)