# Polar Curve Area

• Mar 21st 2008, 04:42 PM
thegame189
Polar Curve Area
Find the area inside both r=5sin(2theta) and r=5cos(2theta).

I have NO idea how to do it. My book has no examples and I couldn't find anything online.

I ended up trying to do the integral of (5sin(2theta))^2, evaluated from 0 to Pi/8 + the integral of (5cos(2theta))^2 evaluated from 0 to Pi/8 and then multiplying the whole thing by 8 because of symmetry. I don't believe that to be correct.

Any explanations would be great!
• Mar 21st 2008, 05:34 PM
galactus
One point the curves intersect in at Pi/8. If we find the area inside

$5sin(2{\theta})$ from 0 to Pi/8 and multiply by 16, we can find the area of all

the regions inside the two functions.

$8\int_{0}^{\frac{\pi}{8}}\left[5sin(2{\theta})\right]^{2}d{\theta}$
• Mar 21st 2008, 05:49 PM
thegame189
First, I don't understand why you disregard the 5cos(2theta) completely. Why do you only use the sin equation?

I also don't understand why you multiply by 16. There are only 8 sections that are in BOTH graphs, so wouldn't you multiply by 8?

Edit: Ah, I see you fixed it. I was wondering why it wasn't squared. Thanks for adding a graph, too.