Question

The value of the integral \( \int_C (2xy - x^2) \, dx + (x^2 + y^2) \, dy \) where \( C \) is the boundary of the region enclosed by \( y = x^2 \) and \( y^2 = x \), described in the positive sense, is

• A. \(-2\)
• B. \(0\)
• C. \(-1\)
• D. \(2\)

Hint:

When given a line integral over a closed curve, check if Green’s theorem can be used to convert to a simpler double integral.

Answer 1

The Correct Option is B

The given problem is to evaluate the line integral \( \int_C (2xy - x^2) \, dx + (x^2 + y^2) \, dy \), where \( C \) is the boundary of the region enclosed by \( y = x^2 \) and \( y^2 = x \). To solve for the value of the integral, we can apply Green's Theorem, which relates a line integral around a simple closed curve to a double integral over the plane region it encloses. Green's Theorem states: \[\int_C M \, dx + N \, dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA\] Here, \( M = 2xy - x^2 \) and \( N = x^2 + y^2 \). Thus, the partial derivatives are:

• \(\frac{\partial N}{\partial x} = 2x\)
• \(\frac{\partial M}{\partial y} = 2x\)

Calculating the expression under the double integral, we have: \[\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 2x - 2x = 0\] Since the integrand is zero, the double integral over the region \( R \) is: \[\iint_R 0 \, dA = 0\] Therefore, by Green's Theorem, the value of the line integral is also \( 0 \).