# Thread: area of the region btwn 2 curves - polar coordinates

1. ## area of the region btwn 2 curves - polar coordinates

HI

So the two curves are r = sin x and r = cos x

The intersecting point is pi/4 which makes r = 1/sq rt 2

now I know A = 1/2 integral sign (S) a? b? r^2 dx

The question is what is a and b equal to? I am thinking that b could equal pi/4 and then I can multiply by 2 to get the full area but I don't know how to figure out what a would

hopefully I have given you all the info you need to help me.

Thanks

Calculus beginner

2. Originally Posted by calcbeg
HI

So the two curves are r = sin x and r = cos x

The intersecting point is pi/4 which makes r = 1/sq rt 2

now I know A = 1/2 integral sign (S) a? b? r^2 dx

The question is what is a and b equal to? I am thinking that b could equal pi/4 and then I can multiply by 2 to get the full area but I don't know how to figure out what a would

hopefully I have given you all the info you need to help me.

Thanks

Calculus beginner
note the symmetry in the graph ...

$A = \int_0^{\frac{\pi}{4}} \sin^2{\theta} \, d\theta$

or ...

$A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta$

3. Now I actually figured out the pi/4 part while I was waiting for someone to guide me but I don't know how you picked the other boundary.

I was going to use a = c for one b = c for the other and then subtract the area of one from the area of the second to get the overlap..... but obviously I am missing the obvious.

By the way Skeeter you are my hero in this calculus world. My husband suggests I find out what you would like for Christmas.

Thanks

Calculus Beginner