Question

If \( \hat{F}(x, y) = (3x - 8y) \hat{i} + (4y - 6xy) \hat{j} \) for \( (x, y) \in \mathbb{R}^2 \), then \( \oint_C \vec{F} \cdot d \vec{r} \), where \( C \) is the boundary of the triangular region bounded by the lines \( x = 0 \), \( y = 0 \), and \( x + y = 1 \) oriented in the anti-clockwise direction, is
 

• A. \( \frac{5}{2} \)
• B. 3
• C. 4
• D. 5

Hint:

Green's Theorem simplifies the evaluation of line integrals by converting them into double integrals over the region enclosed by the curve.

Answer 1

The Correct Option is B

Step 1: Applying Green's Theorem. 
We use Green's Theorem to convert the line integral into a double integral over the region \( R \) enclosed by the curve \( C \): \[ \oint_C \hat{F} \cdot d\hat{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] where \( \hat{F} = P \hat{i} + Q \hat{j} \). Here, \( P = 3x - 8y \) and \( Q = 4y - 6xy \). 
 

Step 2: Computing the Partial Derivatives. 
We compute: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (4y - 6xy) = -6y, \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (3x - 8y) = -8 \] So, the integrand becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -6y + 8 = 8 - 6y \]

Step 3: Setting up the Integral. 
The region \( R \) is a right triangle with vertices at \( (0, 0) \), \( (1, 0) \), and \( (0, 1) \). We set up the double integral over this triangular region: \[ \iint_R (8 - 6y) \, dA \] The limits of integration are \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 - x \). So, the integral becomes: \[ \int_0^1 \int_0^{1-x} (8 - 6y) \, dy \, dx \]

Step 4: Solving the Integral. 
First, integrate with respect to \( y \): \[ \int_0^{1-x} (8 - 6y) \, dy = 8y - 3y^2 \Big|_0^{1-x} = 8(1-x) - 3(1-x)^2 \] Now, integrate with respect to \( x \): \[ \int_0^1 \left[ 8(1-x) - 3(1-x)^2 \right] dx = 3\]

Step 5: Conclusion. 
The correct answer is (B) 3