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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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Um, you sure that $\sin$ in the last integral is supposed to be without any absolute value delimiters? Also, $\cos\cdot\sec\equiv1$ identically... (Other than these two points I haven't really read your question.) – anon Oct 31 '11 at 9:43
Yup, you're right, the sin t is abs value, how does it change my answer?? – adsisco Oct 31 '11 at 11:11
@user10676 I think my ds is right... – adsisco Oct 31 '11 at 11:11
up vote 5 down vote accepted

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. – adsisco Oct 31 '11 at 11:45

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