Integral calculus: plane areas in polar coordinates

[delapcsoncruz]

what is the area inside the graph of r=2sinθ and outside the graph of r=sinθ+cosθ?

so i compute for the values of 'r',... but, i only got one intersection point which is (45°, 1.41).
there must be two intersection points right? but I've only got one. what shall i do?
i cannot compute for the area of the said region because I've only got one limit which is ∏/4.. what shall i do?

please help..
thanks a lot..

 

[tiny-tim]

hi delapcsoncruz!

useful tip: multiply both sides by r, then convert to cartesian coordinates …

that (amost) immediately gives you the positions of these circles

 

[Office_Shredder]

Alternatively not that 5pi/4 is also an intersection point

 

[delapcsoncruz]

can you please give me of an example using your said useful tip.. please..

 

[delapcsoncruz]

Alternatively not that 5pi/4 is also an intersection point

the value of 'r' in 5pi/4 is -1.41 , so that is also equal to pi/4 which r is 1.41

 

[delapcsoncruz]

can you give me an example of your useful tip.. please..

 

[tiny-tim]

try it with r = 2sinθ …

what do you get? ​

 

[delapcsoncruz]

r^2=2rsin(theta)

 

[tiny-tim]

now convert to cartesian (x and y)