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 $r=\sin\theta$
$0\leq r \leq\sin\theta$
$0\leq\theta\leq\Pi$Now 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 $r=\sin\theta$
$0\leq r \leq\sin\theta$
$0\leq\theta\leq\Pi$Now 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...