Question

If \( \vec{F}(x, y, z) = 3x^2y\,\hat{i} + 5y^2z\,\hat{j} - 8xyz\,\hat{k} \) is a continuously differentiable vector field, then the curl of \( \vec{F} \) at (1,1,1) is ...............

• A. $-13\hat{i} - 8\hat{j} - 3\hat{k}$
• B. $-13\hat{i} + 8\hat{j} + 3\hat{k}$
• C. $-13\hat{i} + 8\hat{j} - 3\hat{k}$
• D. $13\hat{i} + 8\hat{j} - 3\hat{k}$

Hint:

Use the determinant form of curl and compute partial derivatives carefully before substituting the point.

Answer 1

The Correct Option is C

We are given a vector field:
\( \vec{F}(x, y, z) = 3x^2y\,\hat{i} + 5y^2z\,\hat{j} - 8xyz\,\hat{k} \)

Let \( F_1 = 3x^2y \), \( F_2 = 5y^2z \), \( F_3 = -8xyz \)

The curl of a vector field is defined as:
\[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ F_1 & F_2 & F_3 \end{vmatrix} \]
Compute each term:
\( \frac{\partial F_3}{\partial y} = -8xz \)
\( \frac{\partial F_2}{\partial z} = 5y^2 \)
\( \Rightarrow \hat{i} \) component = \( -8xz - 5y^2 \)

\( \frac{\partial F_3}{\partial x} = -8yz \)
\( \frac{\partial F_1}{\partial z} = 0 \)
\( \Rightarrow \hat{j} \) component = \( 0 - (-8yz) = 8yz \)

\( \frac{\partial F_2}{\partial x} = 0 \)
\( \frac{\partial F_1}{\partial y} = 3x^2 \)
\( \Rightarrow \hat{k} \) component = \( 0 - 3x^2 = -3x^2 \)

Evaluate at \( (x, y, z) = (1, 1, 1) \):
\( \hat{i} \): \( -8(1)(1) - 5(1)^2 = -8 - 5 = -13 \)
\( \hat{j} \): \( 8(1)(1) = 8 \)
\( \hat{k} \): \( -3(1)^2 = -3 \)

So,
\( \nabla \times \vec{F} = -13\hat{i} + 8\hat{j} - 3\hat{k} \)