Convert this integral to polar coordinates

Question

$$\int_{0}^{2}\int_{0}^{\sqrt{2y-y^2}}f\text{ dx dy}$$

Now we see that that $x=\sqrt{2y-y^2}$, and doing a little rearranging gives us that $x^2+y^2-2y=0$, which turns into

$x^2+(y-1)^2=1$

The bounded region has this shape:

Clearly we can see that $0\leq \theta \leq \dfrac{\pi}{2}$, but for $r$, here is my confusion.

We see that $0\leq r\leq 2$ (from the picture), but we also know that $x^2+y^2=2y$, which means $r^2=2r\sin\theta\to r=2\sin\theta$, and therefore $0\leq r\leq 2\sin\theta$

So which is correct? They both give different evaluations on the integral (I just plugged in $f=1$).

Essentially my question is, how does one find $r$? Sometimes I see $r$ expressed as functions of $\theta$, other times I just see $r$ between two constants (usually $0$ and $1$). What is the correct answer here, and why?

Answer 1

$r :0 \to 2\sin\theta$

$\theta: 0 \to \frac{\pi}{2}$

$dxdy=rdrd\theta$

Answer 2

$$0 \leq r \le 2 \sin \theta$$

Draw a line from the origin to a point on the boundary. It doesn't always end on $r=2$ but it ends on $r=2\sin \theta$.

$$\int_0^\frac{\pi}2 \int_0^{2\sin \theta} fr\,\,dr d\theta$$

Remark:

If you substitute in $f=1$, you should obtain the area of semi-circle to check whether what you are doing is correct.

Answer 3

Your confusion comes from your use of $r$ before fully defining it. In this case, you're trying to use the parametrization of $x, y$ as $r, \theta$:

$$x = r\cos\theta\\y = r\sin\theta.$$

Everything comes from this. You define $r$ and $\theta$ by what they are necessarily because of this equation. The bounds on $r$ with this parameterization are $0<r<2\sin\theta$.

If you wanted $r$ to be between two constants, you would need to define $r$ and $\theta$ in a different way, with something like:

$$x = r_1\cos\theta_1\\y = 1+r_1\sin\theta_1$$

You might see that using this parameterization of your shape yields bounds:

$$0<r_1<1\\-\frac\pi2<\theta_1<\frac\pi2.$$

Does this help clarify what is going on with this?