I'm looking through a triple integral example provided by my university, some explanation would be appreciated!
Find the volume between the paraboloid$\ z=x^2+y^2 $ and the plane$\ z=2y $
A visual representation is also provided.
The solution calls for cylindrical coordinates
$\ |R|= \iiint dV $
Bounding$\ z $ coordinates of$\ R $ by
$\ z_{paraboloid} \le z \le z_{plane} $
$\ x^2+y^2 \le 2y \Rightarrow x^2 + (y-1)^2 = 1 $
This gives the shadow of$\ R $ on the$\ xy $ plane as a circle of radius$\ r=1 $ centred at$\ (x,y)=(0,1) $. Transforming the equation of this circle to polar coordinates yields
$\ r=2\sin \theta $
Looking at the shadow of$\ R $ on the$\ xy $ plane,$\ r:0 \rightarrow 2 \sin \theta$ (from zero to any point on the boundary of the disk and$\ \theta:0 \rightarrow \pi $.
My issue here is, how are the limits of$\ \theta $ found? I can follow the solution up until this point; any help?
The final integral is the written as
$\ |R|=\int_0^{\pi}\int_0^{2\sin\theta}\int_{r^2}^{2r\sin\theta}rdzdrd\theta $
Hint:
As you noted the equation of the circle $x^2+(y-1)^2=1$ in polar coordinates is $r=2\sin \theta$, but with $0\le \theta \le \pi$ because $r$ must be $\ge 0$.