0

Using Green theorem I need to calculate the area bounded with $(x+y)^4=x^2y$. First(after converting to polar coordinates) i get $x=\cos^6\phi\sin^2\phi$ and $y=\cos^4\phi\sin^4\phi$. And after i plug that into Green's formula i get $\int\limits_{0}^{2\pi} \cos^9\phi\sin^5\phi d\phi$ and that is $0$. But when i set bounds for $\phi$ to be $0, \frac{\pi}{2}$ i get correct solution. Why we have $\frac{\pi}{2}$ for $\phi$ when we don't have any restrictions?

Trevor
  • 533
  • 1
    Because what you are integrating is odd so you get negative values in second and fourth quadrants. As you need to find area, you should do $\int_{0}^{2\pi} |\cos^9\phi\sin^5\phi)| d\phi$ – Math Lover Dec 27 '20 at 13:48
  • 1
    The loop doesn't exist in all quadrants. For example the first equation implies $y$ must always be positive. – Ninad Munshi Dec 27 '20 at 13:49

1 Answers1

2

The closed curve is drawn once as $\phi$ goes from $0$ to $\pi/2$, because both the functions for $x$ and $y$ have period $\pi/2$. Then as $\phi$ goes from $\pi/2$ to $\pi$ the same curve is traced backward. So integrating from $0$ to $\pi$ should give $0$. This act is repeated from $\pi$ to $2\pi$.

B. Goddard
  • 1