Converting rectangular coordinates to cylindrical coordinates and then integrating $$\int_0^2 \int_9^\sqrt{2x-x^2} (xy) \ dy\ dx$$ 
I have to solve this problem by converting from rectangular coordinates to cylindrical coordinates then integrate it.
I know that $$\ r^2 = x^2 + y^2 $$
 $$\ x = rcos(θ) $$  $$\ y = rsin(θ) $$
I just am confused on what to do next. Does the integral change to:
$$\int_0^2 \int_9^\sqrt{2x-x^2} (rcos(θ)rsin(θ)) \ dr\ dθ$$ ?
 A: It is not suitable to sovle it by your method. Just do it dicretly.
$$\int_0^2 \int_9^\sqrt{2x-x^2} (xy) \ dy\ dx = \int_0^2 x \int_9^\sqrt{2x-x^2} y\ dy\ dx
$$ 
Can you proceed from here?
A: The graph of $y = \sqrt{2x-x^2}$ looks like this:

So one thing to do is to find the formula $r = f(\theta)$
that produces the same shape in polar coordinates.
This will tell you what to put as one end of the integral over $r$.
What you put as the other end of the integral depends on how much you
trust your professor's proofreading skills.
If he really, really meant the $9$ in
$$\int_0^2 \int_9^\sqrt{2x-x^2} (xy) \,dy\,dx,$$
then you're really being asked to find
$$-\int_0^2 \int_\sqrt{2x-x^2}^9 (xy) \,dy\,dx,$$
and the region to integrate over is the (mostly) rectangular figure
bounded by $x=0$ on the left, $x=2$ on the right,
$y=9$ on top, and this semicircle on the bottom.
You can do the whole thing in polar coordinates, but it's really ugly:
you need to break it into two parts, one for $0\leq \theta\leq \arctan \frac92$
(covering everything under the radial line that passes through
the upper right corner of your region, $(2,9)$)
and a second part for $\arctan \frac92 \leq \theta\leq \frac\pi2.$
On the first integral $r$ runs from a point on the semicircle to a point
on the line $x=2$;
on the second integral $r$ runs from a point on the semicircle to a point
on the line $y=9$.
That is, you're being asked to integrate a rectangle in polar coordinates,
while taking a semicircular chunk out of the bottom.
Alternatively, you can do the integral $\int_0^2 \int_0^9 xy\;dx\,dy$
in rectangular coordinates, then subtract the integral of
$\int_0^2 \int_0^\sqrt{2x-x^2} xy \;dy\,dx,$
that is, you remove the semicircle from the bottom of the rectangle.
The integral $\int_0^2 \int_0^\sqrt{2x-x^2} xy \;dy\,dx,$
looks fairly reasonable to approach in polar coordinates;
at each value of $\theta$, $r$ ranges from $0$ to the 
value of $r$ on the semicircle.
If it was just a typo then you just need to do the integral
$\int_0^2 \int_0^\sqrt{2x-x^2} xy \;dy\,dx$
in polar coordinates.
A handy fact to remember is that the graph of $r = \cos\theta$
in polar coordinates is a circle of diameter $1$ passing through the origin;
by taking $r = 2\cos\theta$ you double the size of that circle and it
now coincides with your desired semicircle.
Just remember to integrate over a large enough range of $\theta$
so that you reach every point on the semicircle, but no farther.
A: $$\int_{0}^{\pi} \int_{0}^{2\cos(\Theta)} rsin(\Theta) rcos(\Theta)\ r \ dr  d\Theta$$
this is how you would do it basically
