Question

Let $\vec{f} = x^2 y z \hat{i} + x y^2 z \hat{j} + x y z^2 \hat{k}$ be a vector field. If $\vec{F}(x,y,z) = \text{curl }\vec{f}$, then the vector $\vec{F}(1,-2,1)$ is equal to _______}

• A. $-3\hat{i} + 4\hat{j} + 3\hat{k}$
• B. $-3\hat{i} + 3\hat{k}$
• C. $3\hat{i} - 3\hat{k}$
• D. $-3\hat{i} - 4\hat{j} + 3\hat{k}$

Hint:

Double check signs when expanding the cross product determinant.
The middle term ($\hat{j}$) has a negative sign in the expansion: $-\hat{j}(\frac{\partial f_z}{\partial x} - \frac{\partial f_x}{\partial z})$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question requires computing the curl of a given three-dimensional vector field $\vec{f}$ and evaluating the resulting curl vector field $\vec{F}$ at the point $(1, -2, 1)$.

Step 2: Key Formula or Approach:
The curl of a vector field $\vec{f} = f_x \hat{i} + f_y \hat{j} + f_z \hat{k}$ is defined as:
\[ \text{curl }\vec{f} = \nabla \times \vec{f} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} f_x & f_y & f_z \end{vmatrix} \] Expanding this determinant yields:
\[ \text{curl }\vec{f} = \left( \frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z} \right) \hat{i} - \left( \frac{\partial f_z}{\partial x} - \frac{\partial f_x}{\partial z} \right) \hat{j} + \left( \frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y} \right) \hat{k} \]

Step 3: Detailed Explanation:

• Identify the components of $\vec{f}$:
\[ f_x = x^2 y z, \quad f_y = x y^2 z, \quad f_z = x y z^2 \]
• Calculate each component of the curl:

• $\hat{i}$-component:
\[ \frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z} = \frac{\partial}{\partial y}(x y z^2) - \frac{\partial}{\partial z}(x y^2 z) = x z^2 - x y^2 \]
• $\hat{j}$-component:
\[ -\left( \frac{\partial f_z}{\partial x} - \frac{\partial f_x}{\partial z} \right) = -\left[ \frac{\partial}{\partial x}(x y z^2) - \frac{\partial}{\partial z}(x^2 y z) \right] \] \[ = -(y z^2 - x^2 y) = x^2 y - y z^2 \]
• $\hat{k}$-component:
\[ \frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y} = \frac{\partial}{\partial x}(x y^2 z) - \frac{\partial}{\partial y}(x^2 y z) = y^2 z - x^2 z \]
• Combine these parts to write the curl vector field $\vec{F}(x,y,z)$:
\[ \vec{F}(x,y,z) = (x z^2 - x y^2) \hat{i} + (x^2 y - y z^2) \hat{j} + (y^2 z - x^2 z) \hat{k} \]
• Now, substitute the coordinates of the point $(1, -2, 1)$, i.e., $x = 1$, $y = -2$, and $z = 1$:

• $\hat{i}$-component:
\[ (1)(1)^2 - (1)(-2)^2 = 1 - 4 = -3 \]
• $\hat{j}$-component:
\[ (1)^2(-2) - (-2)(1)^2 = -2 - (-2) = 0 \]
• $\hat{k}$-component:
\[ (-2)^2(1) - (1)^2(1) = 4 - 1 = 3 \]
• This gives the final vector:
\[ \vec{F}(1, -2, 1) = -3\hat{i} + 3\hat{k} \]

Step 4: Final Answer:
The vector is $-3\hat{i} + 3\hat{k}$, which corresponds to option (B).