Finding Limits of r^2=4sin2ø (Polar Coordinates)

**Audioslavery** · April 21st 2010, 04:08 PM

I have to find the area of this region, but I'm not quite sure what my limits are for it.

Thanks for your help!

r^2=4sin2ø

**TKHunny** · April 21st 2010, 05:20 PM

Plot points until you see it. Watch out for multiple laps and duplicity and crossing over itself.

$\theta = 0$

Please pardon the 2nd grade algebra lesson. It is more confusing than may be realized. It helps to back up to fundamentals and go one tiny step at a time.

$r^{2} = 4\sin(2*0) \implies r^{2} = 4\sin(0) \implies r^{2} = 4*0 \implies r^{2} = 0 \implies r = 0$

I was worried we were going to get two values. This is an excellent warning that we WILL for future values.

So far, we have (0,0)

$\theta = \frac{\pi}{6}$

$r^{2} = 4\sin(2*\frac{\pi}{6}) \implies r^{2} = 4\sin(\frac{\pi}{3}) \implies r^{2} = 4*\frac{\sqrt{3}}{2}$
$\implies r^{2} = 2\sqrt{3} \implies r = \pm \sqrt{2}\sqrt[4]{3}$

...or about 1.861

So we jumped rather far from the Origin and spawned two parts of the graph. This is VERY important. If nothing else, it might mean that limits of $[0,\pi]$ will cover the whole figure, rather than requiring $[0,2\pi]$

This same effort with maybe $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , and $\frac{\pi}{2}$ will show that this is all it takes to trace the ENTIRE curve. A little thought may suggest that continuing will retrace the curve another time, perhaps a little differently, but trace nonetheless. In this case, it seems that the result goes off the Real plane and you do not get a retrace on $[\frac{\pi}{2},\pi]$ , but it comes back for another tracing after that.

With a little practice, you WILL get better at it.

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