Green's theorem in the plane states that for a positively oriented, piecewise-smooth, simple closed curve $C$ bounding a region $R$:
$\oint_C (M dx + N dy) = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA$
Evaluating $\oint_C (x^2 y \, dx + x y^2 \, dy)$ over any circle centered at the origin gives:
Options:
(A) 0
(B) $\pi$
(C) $2\pi$
(D) $4\pi$
Answer Key:
A
Step-by-Step Solution & Explanation
Identify $M(x, y) = x^2 y$ and $N(x, y) = x y^2$.
Compute partial derivatives:
$\frac{\partial N}{\partial x} = y^2, \quad \frac{\partial M}{\partial y} = x^2$
By Green's theorem:
$\oint_C (M dx + N dy) = \iint_R (y^2 - x^2) \, dA$
By symmetry over any region $R$ that is a disk centered at the origin (or in polar coordinates where $x = r\cos\theta, y = r\sin\theta$):
$\int_0^{2\pi} (\sin^2\theta - \cos^2\theta) d\theta = \int_0^{2\pi} -\cos(2\theta) d\theta = 0$
Hence, the integral is strictly $0$.