Answer to Question #204889 in Physics for DEBANKUR BISWAS

Question

Using Green's theorem evaluate the integral: ∫C (y³dx + 3x²dy) where C is the contour along the circle x² + y² = 1 taken counter clockwise.

Answer 1

According to Green's theorem, $\int_C P dx + Q dy = \iint_D \left(\frac{ \partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$.

For given vector field, $\frac{ \partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6 x - 3 y^2$, hence we need to evaluate surface integral$\iint (6 x - 3 y^2) dx dy$ over the circle of radius 1, centered at the origin.

It is convenient to do it in polar coordinates, where surface element is $dS = r dr d\varphi$, and $0\leq r \leq 1$, $0\leq \varphi \leq 2 \pi$:

$\int_0^{2\pi} d\varphi \int_0^1 r (6 r \cos \varphi - 3 r^2 \sin^2 \varphi) = \int_0^{2\pi}(\frac{6}{3}\cos \varphi - \frac{3}{4} \sin^2 \varphi)d \varphi =$

$\left[ 2 \sin \varphi - \frac{3}{4}(\frac{\varphi}{2} - \frac{1}{4} \sin 2 \varphi)\right]_0^{2 \pi} = -\frac{3}{4}\pi$.