Question

If C is a triangle with vertices (0,0), (1,0) and (1,1) which are oriented counter clockwise, then \( \oint_C 2xydx + (x^2+2x)dy \) is equal to:

• A. \( \frac{1}{2} \)
• B. 1
• C. \( \frac{3}{2} \)
• D. 2

Hint:

For line integrals over closed loops in the plane, always consider Green's Theorem first. It's often much simpler than parameterizing each segment of the path, especially if the term \( (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \) simplifies to a constant or a simple function.

Answer 1

The Correct Option is B

Step 1: Problem Overview:
The problem requires evaluating a line integral over a closed triangular path. Green's Theorem is ideal for this, transforming the line integral into a double integral over the enclosed region.

Step 2: Theorem and Setup:
Green's Theorem: \( \oint_C P dx + Q dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \).
Identify \( P \) and \( Q \) from the integral:- \( P(x,y) = 2xy \)- \( Q(x,y) = x^2 + 2x \)The region \( R \) is a triangle with vertices (0,0), (1,0), and (1,1).

Step 3: Calculation:
Compute partial derivatives:\[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 + 2x) = 2x + 2 \]\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2xy) = 2x \]Calculate the difference:\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (2x + 2) - 2x = 2 \]Apply Green's Theorem:\[ \oint_C P dx + Q dy = \iint_R 2 \, dA = 2 \iint_R dA \]The double integral \( \iint_R dA \) equals the area of region \(R\), a right triangle with base 1 and height 1.
Triangle area:\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \]Substitute the area:\[ \oint_C P dx + Q dy = 2 \times (\text{Area of R}) = 2 \times \frac{1}{2} = 1 \]
Step 4: Solution:
The line integral's value is 1.