<s>[INST] A double integral into polar coordinates I have the double integral
$$\int^{10}_0 \int^0_{-\sqrt{10y-y^2}} \sqrt{x^2+y^2} \,dx\,dy$$
And I am asked to evaluate this by changing to polar coordinates.
 [/INST] Complete the square:
$$
10y-y^2 = 25 - (5-y)^2,
$$
so the graph of $x = -\sqrt{10y-y^2} = - \sqrt{5^2 - (5-y)^2}$ is the left half of the circle $x^2 + (5-y)^2 = 5^2.$
Exercises with polar coordinates will have shown you that $$\tag 1 r = 5\sin\theta$$ is that circle. If you multiply both sides of $(1)$ by $r,$ you get $$ r^2 = 5r\sin\theta $$ which becomes $$ x^2+y^2 = 5y $$ and by completing the square, then becomes $$ x^2 + (y-5)^2 = 5^2. $$
Thus you have
$$
\int_{\pi/2}^\pi \int_0^{5\sin\theta} r\, dr\,d\theta.
$$
</s>