Question

The line integral of the vector field \( \vec{F} = (zx, xy, yz) \) along the boundary of the triangle with vertices (1,0,0), (0,1,0), (0,0,1) anticlockwise, when viewed from the point (2,2,2) is

• A. \( -\frac{1}{2} \)
• B. \( -2 \)
• C. \( \frac{1}{2} \)
• D. \( 2 \)

Hint:

Stokes' theorem simplifies line integrals over closed curves: \( \oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \). Ensure the orientation matches the direction of traversal.

Answer 1

The Correct Option is C

Step 1: Understand the problem and apply Stokes' theorem.
We need to compute the line integral \( \oint_C \vec{F} \cdot d\vec{r} \), where \( \vec{F} = (zx, xy, yz) \), and \( C \) is the boundary of the triangle with vertices \( (1,0,0) \), \( (0,1,0) \), and \( (0,0,1) \), traversed anticlockwise when viewed from \( (2,2,2) \). Stokes' theorem states: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}, \] where \( S \) is the surface enclosed by \( C \), and \( d\vec{S} \) is the surface normal vector consistent with the direction of traversal. Step 2: Determine the orientation of the triangle.
The triangle has vertices \( A(1,0,0) \), \( B(0,1,0) \), and \( D(0,0,1) \). The plane equation of the triangle is found using the points: \[ x + y + z = 1 \quad (\text{since } 1x + 0y + 0z = 1 \text{ at } (1,0,0), \text{ etc.}). \] The normal to the plane is \( \nabla (x + y + z - 1) = (1,1,1) \). When viewed from \( (2,2,2) \), the point is above the plane (\( 2+2+2=6>1 \)), so anticlockwise traversal corresponds to the normal \( (1,1,1) \). The path \( A \to B \to D \to A \) is anticlockwise when viewed from above. Step 3: Compute the curl of the vector field.
Given \( \vec{F} = (zx, xy, yz) \), compute \( \nabla \times \vec{F} \): \[ \nabla \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}
zx & xy & yz \end{vmatrix}, \] \[ = \mathbf{i} \left( \frac{\partial (yz)}{\partial y} - \frac{\partial (xy)}{\partial z} \right) - \mathbf{j} \left( \frac{\partial (yz)}{\partial x} - \frac{\partial (zx)}{\partial z} \right) + \mathbf{k} \left( \frac{\partial (xy)}{\partial x} - \frac{\partial (zx)}{\partial y} \right), \] \[ = \mathbf{i} (z - 0) - \mathbf{j} (0 - x) + \mathbf{k} (y - 0) = (z, x, y). \] So, \( \nabla \times \vec{F} = (z, x, y) \). Step 4: Set up the surface integral.
The surface \( S \) is the triangle in the plane \( x + y + z = 1 \). The normal \( d\vec{S} = (1,1,1) \, dA \), where \( dA \) is the area element in the projection onto the \( xy \)-plane. Project the triangle onto the \( xy \)-plane (\( z = 0 \)):
Vertices: \( (1,0,0) \to (1,0) \), \( (0,1,0) \to (0,1) \), \( (0,0,1) \to (0,0) \).
The region is the triangle bounded by \( (1,0) \), \( (0,1) \), \( (0,0) \), with equation \( y = 1 - x \).
On the surface, \( z = 1 - x - y \), so \( \nabla \times \vec{F} = (z, x, y) = (1 - x - y, x, y) \). The dot product is: \[ (\nabla \times \vec{F}) \cdot d\vec{S} = (1 - x - y, x, y) \cdot (1,1,1) \, dA = (1 - x - y) + x + y = 1. \] Thus, the surface integral becomes: \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_D 1 \, dA, \] where \( D \) is the projected triangle in the \( xy \)-plane. Step 5: Compute the area of the projected triangle.
The triangle in the \( xy \)-plane has vertices \( (0,0) \), \( (1,0) \), \( (0,1) \). The area is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}. \] So: \[ \iint_D 1 \, dA = \text{Area of } D = \frac{1}{2}. \] By Stokes' theorem: \[ \oint_C \vec{F} \cdot d\vec{r} = \frac{1}{2}. \] Step 6: Evaluate the options.
(1) \( -\frac{1}{2} \): Incorrect, as the integral is positive. Incorrect.
(2) \( -2 \): Incorrect, as the integral is positive and smaller. Incorrect.
(3) \( \frac{1}{2} \): Correct, matches the computed value. Correct.
(4) \( 2 \): Incorrect, as the integral is smaller. Incorrect.
Step 7: Select the correct answer.
The line integral is \( \frac{1}{2} \), matching option (3).