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Over a curve $C$ given by $(x^2+y^2)^2=30^2(x^2-y^2)$, What is $ \oint\limits_C |y|\,\mathrm ds. $

I've tried working on it but I couldn't get the solution.

Here's how I did it:

Using polar coordinates $ \begin{cases} x(t) &= 30 \sqrt{\cos 2t}\cdot\cos t \\ y(t) &= 30 \sqrt{\cos 2t} \cdot\sin t \end{cases} $ thereafter, $\mathrm ds = 30 \sqrt{ \sec 2t}\cdot \mathrm dt$.

Finally, integral over $ \oint\limits_C |y|\,ds = 2 \cdot 30^2\int\limits_{-\pi/4}^{\pi/4}\sqrt{\cos 2t} \cdot\sqrt{\sec 2t} \sin t\, \mathrm dt = 0. $

I've spent many hours on this problem already. Will someone be kind enough to please help me? Thanks.

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    @user10676 I think my ds is right...2011-10-31

1 Answers 1

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Your parametrization of $C$ is correct, but note that only for $t$ in the intervals $\bigl[-{\pi\over4},{\pi\over4}\bigr]$ and $\bigl[{3\pi\over4},{5\pi\over4}\bigr]$ we actually get points of $C$. As $r(t)=30\sqrt{\cos(2t)}$ one has s'(t)=\sqrt{r^2(t)+r'^2(t)}={30\over\sqrt{\cos(2t)}}={30^2\over r(t)}\ , in accordance with your ${\rm d}s$. It follows that $|y| ds=r(t)\ |\sin t|\ {30^2\over r(t)}=30^2|\sin t|\ dt\ ,$ so that we finally obtain $\int_C|y|\ {\rm d}s=2\cdot30^2\cdot\int_{-\pi/4}^{\pi/4}|\sin t|\ dt=1800\bigl(2-\sqrt{2}\bigr)\ .$

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    Hey! thanks for the great answer! heres an up vote.2011-10-31