I am given an equation of a cylinder $z = x^2+(y-1)^2=1$
When graphed, the circle is centered at (0,1). When taking the integral of this cylinder, I need to find the value for $r$, which I can do by rewriting the equation as follows:
$x^2+y^2-2y+1=1 \Longrightarrow x^2+y^2 = 2y$
Then, converting to polar notation, it turns into
$r^2 = 2sin(\theta)$
Now, when taking the integral, I've been instructed to integrate from $0$ to $r$, so if $r^2 = 16$ I would integrate from $0$ to $4$.
However, in this case the solution says that I should integrate from $0$ to $2sin(\theta)$
I understand where the $2sin(\theta)$ came from, but shouldn't it be $sqrt(2sin(\theta))$ instead? Why am I not taking the root if it is the value of $r^2$ like I do when it has a constant radius?
You're really close!
It is true that $$x^2+y^2-2y+1=1 \Longrightarrow x^2+y^2 = 2y$$
As you say, polar coordinates give us
$$r^2 = x^2 + y^2$$
but polar coordinates give $$ y = \mathbf r \sin(\theta) $$
not $$y=\sin(\theta) $$
So we get $$r^2 = 2\mathbf r \sin(\theta)$$
and by dividing by $r$ $$r = 2 \sin(\theta)$$