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 [itex]r=\cos\theta[/itex]
[itex]0\leq r \leq\cos\theta[/itex]

[itex]-\frac{\Pi}{2}\leq\theta\leq\frac{\Pi}{2}[/itex]

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

For r we must find which function is the smallest and use that but [itex]\sin\theta[/itex] is greater than [itex]\cos\theta[/itex] 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 [itex]r=\cos\theta[/itex]
[itex]0\leq r \leq\cos\theta[/itex]

[itex]-\frac{\Pi}{2}\leq\theta\leq\frac{\Pi}{2}[/itex]

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

For r we must find which function is the smallest and use that but [itex]\sin\theta[/itex] is greater than [itex]\cos\theta[/itex] 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 [itex]\theta= -\pi/4[/itex] to [itex]\pi/4[/itex] r goes from 0 to [itex]sin(\theta)[/itex] while from [itex]\theta= \pi/4[/itex] to [itex]3\pi/4[/itex], r goes from 0 to [itex]cos(\theta)[/itex].

 

[mrcleanhands]

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