Answer to Question #95176 in Physics for Manisha Nayak

Question #95176
Using green's theorem evaluate the integral ellipse c (x2ydx-xy2dy) where c is the circle x2 +y2 =4 oriented counter clockwise.
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Expert's answer
2019-09-26T14:48:17-0400

We need to evaluate CFds\oint_C \bold F d \bold s, where F=(P(x,y),Q(x,y))=(x2y,xy2)\bold F = (P(x,y), Q(x,y)) = (x^2 y, -x y^2), and CC is the contour of the circle of radius 22, centered at the origin.

According to Green's theorem, CFds=D(QxPy)dA=(y2x2)dA=(x2+y2)dA\oint_C \bold F d \bold s = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial{y}})d A = \iint (-y^2 - x^2) d A = -\iint (x^2+y^2) dA , where we need to take the surface integral inside the circle.

Let us use polar coordinates: x=rcosθ,y=rsinθ,dA=rdrdθx = r \cos \theta, y = r \sin \theta, d A = r dr d\theta , 0r2,0θ2π0\leq r \leq 2, 0 \leq \theta \leq 2 \pi. Hence, the integrand is 02πdθ02rr2dr=2πr4402=8π-\int_0^{2\pi} d \theta \int_0^2 r \cdot r^2 dr = -2 \pi \frac{r^4}{4}|_0^2 = -8 \pi.


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