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$ \int(x-y)dS $ where $ r^2=a^2\cos(2\phi) $ and $ -\frac{\pi}{4} \leq \phi \leq \frac{\pi}{4}$ Since $ r=a\sqrt{\cos(2\phi)}$, do I convert $ x=r\cos(\phi) $ and $ y=r\sin(\phi) $ into $ x=a\sqrt{\cos(2\phi)}\cos(\phi) $ and $ y=a\sqrt{\cos(2\phi)}\sin(\phi) $ and then calculate $ dS=\sqrt{\left(\frac{dx}{d\phi}\right)^2 + \left(\frac{dy}{d\phi}\right)^2} $ or do I calculate $dS$ by leaving $ x=r\cos(\phi)$ and $y=r\sin(\phi)$?

I'm asking because I seem to get slightly different solutions.

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    @MichaelHardy Done. You're absolutely right about that.2012-03-03

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I'm also learning line integration, but I think I got this one right.

You want to find

$\sqrt {{{\left( {\frac{{dx}}{{d\phi }}} \right)}^2} + {{\left( {\frac{{dy}}{{d\phi }}} \right)}^2}} $

We have that

$\eqalign{ & \frac{{dx}}{{d\phi }} = \frac{{dr}}{{d\phi }}\cos \phi - r\sin \phi \cr & \frac{{dy}}{{d\phi }} = \frac{{dr}}{{d\phi }}\sin \phi + r\cos \phi \cr} $

So we get

$\eqalign{ & {\left( {\frac{{dx}}{{d\phi }}} \right)^2} = {\left( {\frac{{dr}}{{d\phi }}} \right)^2}{\cos ^2}\phi - 2\frac{{dr}}{{d\phi }}r\cos \phi \sin \phi + {r^2}{\sin ^2}\phi \cr & {\left( {\frac{{dy}}{{d\phi }}} \right)^2} = {\left( {\frac{{dr}}{{d\phi }}} \right)^2}{\sin ^2}\phi + 2\frac{{dr}}{{d\phi }}r\cos \phi \sin \phi + {r^2}{\sin ^2}\phi \cr} $

This means that

${\left( {\frac{{dx}}{{d\phi }}} \right)^2} + {\left( {\frac{{dy}}{{d\phi }}} \right)^2} = {\left( {\frac{{dr}}{{d\phi }}} \right)^2} + {r^2}$

(Actually, this is a known result, namely that in polar coordinates one has ds = \sqrt{r^2+r'^2}d\phi

Moving on, we have

$\eqalign{ & {r^2} = {a^2}\cos 2\phi \cr & {\left( {\frac{{dr}}{{d\phi }}} \right)^2} = {a^2}\frac{{\sin 2\phi }}{{\cos 2\phi }}\sin 2\phi \cr} $

So

$\sqrt {{{\left( {\frac{{dr}}{{d\phi }}} \right)}^2} + {r^2}} = \sqrt {\frac{{{a^2}\left( {{{\sin }^2}2\phi + {{\cos }^2}2\phi } \right)}}{{\cos 2\phi }}} = \frac{a}{{\sqrt {\cos 2\phi } }}$

The integral ends up being rather "user friendly"

$\eqalign{ & {a^2}\int\limits_{ - \pi /4}^{\pi /4} {\frac{{\sqrt {\cos 2\phi } \cos \phi - \sqrt {\cos 2\phi } \sin \phi }}{{\sqrt {\cos 2\phi } }}} d\phi \cr & {a^2}\int\limits_{ - \pi /4}^{\pi /4} {\left( {\cos \phi - \sin \phi } \right)} d\phi = {a^2}\sqrt 2 \cr} $