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This one should be a quickie:

if we are calculating $\oint xdy+ydx$ around the curve $x^4 + y^4 = 1$, I can spot quickly that $\oint xdy+ydx = \oint df$ where $f = xy$. How would I then evaluate $\oint df$? Is it simple, or am I better off with a parameterisation approach ($x^2 = \mathrm{cos} \ t$ and so on)?

Poo2uhaha
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    Since you found $f$ then you could just use the fundamental theorem of line integral to show that the line integral is zero. – Jacky Chong Jan 13 '21 at 19:14
  • I had thought something along those lines might occur but it was right at the back of my mind. Thank you. – Poo2uhaha Jan 13 '21 at 19:29

1 Answers1

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Your form $\omega=ydx+xdy$ is defined in $\Omega=\mathbb R^2$, which is contractible, hence it's simply connected (the path $\gamma\equiv x^4+y^4=1$ is closed and continuous).
We can now solve the exercise using the fact that a form in $\Omega$ is closed $\iff$ the form is exact, so we observe that $$\dfrac{\partial}{\partial_y}y=1=\dfrac{\partial}{\partial_x}x\overset{DF}{\implies}\omega\text{ is closed}\implies\omega\text{ is exact.}$$

Vajra
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