Answer to Question #260097 in Calculus for tarie

Question

Use Green’s theorem to calculate the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1. Where F(x, y) =< P, Q >=< y 2 , x 2 >.

Answer 1

\iint_C(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy=\\ \int_{\partial C}(Pdx+Qdy)=[We\space take\space P=- \frac{y}{2},Q=\frac{x}{2}]= \\ [\partial C: x=a\cdot cos(t),y=b\cdot sin(t),0\le t \le 2\pi]=\\ \frac{1}{2}\cdot \int_0^{2\pi}(ab\cdot sin^2(t)+ab\cdot cos^2(t))dt=\frac{1}{2}\cdot ab\cdot \int_0^{2\pi}(sin^2(t)+ cos^2(t))dt=\\ \frac{1}{2}ab\cdot \int _0^{2\pi}dt=ab\pi;\\

From another hand we have

$\iint_C(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy=\\ \iint_C(\frac{\partial (\frac{x}{2})}{\partial x}-\frac{\partial (-\frac{y}{2})}{\partial y})dxdy=\iint_C(\frac{1}{2}+\frac{1}{2})dxdy=\iint_Cdxdy=|C|$

Thus |C|=S(C)=ab$\pi$