1. ## Polar Graph Area

Find the area for one loop of $\displaystyle r^{2} = 64sin2\theta$

I plotted three graphs; $\displaystyle r^{2}$ vs. [tex]\theta]/math], where it looks like a normal sin curve repeating twice between 0 and $\displaystyle 2\pi$ ranging from 64 to -64. Another of r vs. $\displaystyle \theta$ where it looks kind of like two elliptical shapes ranging from -8 to 8, restricted to 0 to $\displaystyle \frac{pi}{2}$ and $\displaystyle \pi$ to $\displaystyle \frac{3\pi}{2}$. Finally, the polar graph on x and y axis, where it looks like a lemniscate reaching a limit of 8 at $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.

I did this:

$\displaystyle A=\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(64sin2\theta) d\theta$

Solving that I got $\displaystyle 8\pi$. I think the answer is $\displaystyle 8\pi - 16$, though. Someone suggested I set the interval to $\displaystyle 2 \int_{0}^{\frac{\pi}{4}}$. I understand what they're doing... integrating to the maximum r and doubling that, but I don't understand why that gives the correct answer and doing it the other way does not.

I think I may have a couple of different computational errors going on, despite trying and retrying the problem, as well as a conceptual block that I'm just not getting.

Any help, much appreciated.

2. Originally Posted by DarrenM
Find the area for one loop of $\displaystyle r^{2} = 64sin2\theta$

I plotted three graphs; $\displaystyle r^{2}$ vs. [tex]\theta]/math], where it looks like a normal sin curve repeating twice between 0 and $\displaystyle 2\pi$ ranging from 64 to -64. Another of r vs. $\displaystyle \theta$ where it looks kind of like two elliptical shapes ranging from -8 to 8, restricted to 0 to $\displaystyle \frac{pi}{2}$ and $\displaystyle \pi$ to $\displaystyle \frac{3\pi}{2}$. Finally, the polar graph on x and y axis, where it looks like a lemniscate reaching a limit of 8 at $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.

I did this:

$\displaystyle A=\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(64sin2\theta) d\theta$
Looks ok to me.

Solving that I got $\displaystyle 8\pi$.
No, I think that's wrong: I get $\displaystyle \int_0^{\pi/4}32\sin(2\theta)\, d\theta = \Big[-16\cos(2\theta)\Big]_{\theta=0}^{\pi/4}=16+16=32$

I think the answer is $\displaystyle 8\pi - 16$, though.
The difference to 32 is so large that it should be possible to see it quite easily in your plot of the graph.

3. Wow. Idiotic mistake. $\displaystyle sin2\theta$ is not even remotely the same as $\displaystyle sin^{2}\theta$. As such, using the double-angle identity would probably be an incredibly stupid thing to do.