Question

Let $\vec{F} = y e^z \hat{i} + x e^z \hat{j} + x y e^z \hat{k}$, then integral of $\vec{F}$ around the boundary of oriented surface $S$ is

• A. 3
• B. 1
• C. 0
• D. 2

Hint:

Whenever asked for a closed loop integral $\oint_C \vec{F} \cdot d\vec{r}$, first check if $\nabla \times \vec{F} = \vec{0}$. If it is irrotational, the integral is $0$ immediately without performing any integration!

Answer 1

The Correct Option is C

Step 1: Concept
By Stokes' Theorem, the line integral of a vector field $\vec{F}$ along a simple closed boundary curve $C = \partial S$ is equal to the surface integral of its curl over $S$: \[ \oint_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS \] If the vector field $\vec{F}$ is conservative (irrotational), then $\nabla \times \vec{F} = \vec{0}$, which implies the loop integral is identically zero.

Step 2: Key Formulas and Approach
Compute $\text{curl}(\vec{F}) = \nabla \times \vec{F}$ using the determinant definition: \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} F_x & F_y & F_z \end{vmatrix} \] where $F_x = y e^z$, $F_y = x e^z$, $F_z = x y e^z$.

Step 3: Step-by-step Explanation

• Evaluate the $\hat{i}$-component of curl: \[ (\nabla \times \vec{F})_x = \frac{\partial}{\partial y}(x y e^z) - \frac{\partial}{\partial z}(x e^z) = x e^z - x e^z = 0 \]
• Evaluate the $\hat{j}$-component of curl: \[ (\nabla \times \vec{F})_y = \frac{\partial}{\partial z}(y e^z) - \frac{\partial}{\partial x}(x y e^z) = y e^z - y e^z = 0 \]
• Evaluate the $\hat{k}$-component of curl: \[ (\nabla \times \vec{F})_z = \frac{\partial}{\partial x}(x e^z) - \frac{\partial}{\partial y}(y e^z) = e^z - e^z = 0 \]
• Therefore, $\nabla \times \vec{F} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$.
• Alternatively, observe that $\vec{F} = \nabla(x y e^z)$ is a conservative gradient field with scalar potential $\phi(x, y, z) = x y e^z$.
• The integral of any conservative field around a closed loop is always 0: \[ \oint_{\partial S} \vec{F} \cdot d\vec{r} = 0 \]

Step 4: Final Answer
The line integral around the boundary of the surface $S$ is 0. Thus, Option (C) is correct.