The region within circles r=cosθ and r=sinθ

[mrcleanhands]

Homework Statement

Use a double integral to find the area of the region enclosed within both circles of r=cosθ and r=sinθ

Homework Equations

The Attempt at a Solution

I begin by finding the region in polar co-ordinates.

For r=\cos\theta
0\leq r \leq\cos\theta

-\frac{\Pi}{2}\leq\theta\leq\frac{\Pi}{2}

For r=\sin\theta
0\leq r \leq\sin\theta
0\leq\theta\leq\PiNow we find the region both of these have in common
which is 0\leq\theta\leq\frac{\Pi}{2}

For r we must find which function is the smallest and use that but \sin\theta is greater than \cos\theta for some portion and smaller for another so not sure what to do here.

 

[LCKurtz]

Homework Statement

Use a double integral to find the area of the region enclosed within both circles of r=cosθ and r=sinθ

Homework Equations

The Attempt at a Solution

I begin by finding the region in polar co-ordinates.

For r=\cos\theta
0\leq r \leq\cos\theta

-\frac{\Pi}{2}\leq\theta\leq\frac{\Pi}{2}

For r=\sin\theta
0\leq r \leq\sin\theta
0\leq\theta\leq\PiNow we find the region both of these have in common
which is 0\leq\theta\leq\frac{\Pi}{2}

For r we must find which function is the smallest and use that but \sin\theta is greater than \cos\theta for some portion and smaller for another so not sure what to do here.

Remember, r goes from r = 0 to r on the outer curve for your region. Do you see that the "outer curve" of your region is a 2 piece formula? You have to set up two integrals, unless you use some shortcut like symmetry.

 

[HallsofIvy]

If you were to look at a graph (these are circles) you would see that from \theta= -\pi/4 to \pi/4 r goes from 0 to sin(\theta) while from \theta= \pi/4 to 3\pi/4, r goes from 0 to cos(\theta).

 

[mrcleanhands]

Ahh ok I see. I wouldn't have been sure that they are symmetrical without first looking at the graph...