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$\oint_{\Gamma}^{} (x^{\frac{4}{3}}+y^{\frac{4}{3}})ds,\Gamma \text{ is } x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ my process is below

$$\left\{\begin{matrix} x=r\text{cos}\theta \\ y=r\text{sin}\theta \end{matrix}\right. \ \theta \in [0,2\pi]$$ $\Rightarrow$ $$I=a^\frac{4}{3}\int_0^{2\pi}(\cos^4\theta+\sin^4\theta)3a\sin\theta\cos\theta d\theta$$ $\Rightarrow$ $$3a^\frac{7}{3}\left(-\int_0^{2\pi}\cos^5d(\cos\theta)+\int_0^{2\pi}\sin^5d(\sin\theta)\right)$$ it is zero, but the answer is $4a^\frac{7}{3}$ The answer uses symmetry to limit the range between $0$ and $\frac{\pi}{2}$. Why is my answer wrong and should use symmetry to limit the range first?

And what kind of problem should I limit the range first?

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  • $\begingroup$ Why did you take the parametrization of the circle? That curve is called astroid en.wikipedia.org/wiki/…. $\endgroup$
    – PNDas
    Feb 14, 2023 at 14:16
  • $\begingroup$ Yes, this problem can also be solved without parametrization, thank you $\endgroup$
    – liyushu
    Feb 14, 2023 at 14:22
  • $\begingroup$ So you solved it? $\endgroup$
    – PNDas
    Feb 14, 2023 at 14:25
  • $\begingroup$ yeah, but I still don't know why my other solution is incorrect $\endgroup$
    – liyushu
    Feb 14, 2023 at 14:32
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    $\begingroup$ I don't remember line integarl much but please tell me if I'm incorrect. You're writing $\int_{\Gamma}f(x,y)\,ds=\int_0^{2\pi}f(a\cos^3t,a\sin^3t) \sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2}\,dt$, right? Then the $3a\cos\theta\sin\theta$ term will be $3a|\cos\theta||\sin\theta|$. So may be the terms will not cancel and you'll not get $0$. $\endgroup$
    – PNDas
    Feb 14, 2023 at 14:49

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