Evaluate iterated integral by changing to polar coordinates $$\int_0^{1/2}\int_0^{\sqrt{1-x^2}}xy\sqrt{x^2+y^2}\,dy\,dx$$
$x^2+y^2=r^2$
$$\int\int_0 r^3\cos\theta \sin\theta|r|\,dr\,d\theta$$
I don't know what $r =$ at line $x = 1/2$. I don't know value of $\theta$. Help me evaluate.
 A: 
The region of integration represented by D can be expressed as $D= D_1 \ U \ D_2$ where $$D_1 = \{ (r, \theta)| 0 \le r \le 1, {\pi\over 3} \le \theta \le {\pi\over 2} \}$$ and $$D_2 = \{ (r, \theta) | 0 \le \theta \le {\pi\over 3}, 0 \le r \le {1\over {2cos \theta}} \}$$ Then you can split the double integral over the two non - intersecting domains and compute the required integral.
A: With this kind of question its good to sketch the region. 
We have $0 \leq y \leq \sqrt{1-x^2}$, and $0 \leq x \leq 1/2$. 
The first chain of inequalities can be re-written as $y\geq 0, y^2 \leq 1 - x^2$ or $1 \geq y^2 + x^2$, $y\geq 0$. This is the upper-half unit disk. Then we have the vertical strip between $0$ and $1/2$ for $x$.
Try to draw a picture of this. You'll see that the region lies in the first quadrant. This gives you a range for $\theta$ between $0$ and $\pi/2$. Then, you have to find an expression for the $r$ that works for $\theta$, in terms of $\theta$.
Wolfram gives this sketch: http://www.wolframalpha.com/input/?i=+0+%3C+x+%3C+1%2F2%2C+x%5E2+%2B+y%5E2+%3C+1%2C+y%3E0
Alternatively, you can try to find $\theta$ in terms of $r$. Again, try to find the range of values that $r$ varies between (you may be able to read it off the diagram, or, you could re-write the inequalities above in terms of $r$ and $\theta$.) Then for each value of $r$, work out the range of $\theta$ that lies within the region.
A: $R \in [0,1]$ because the radious isn't affected by the change of coordinates but you should take some care about $\theta$. 
When $x=\dfrac{1}{2}$, then $y=\dfrac{\sqrt{3}}{2}$ then $\theta=tan^-1(\sqrt{3})=\dfrac{\pi}{3}$ so $\theta \in [0, \dfrac{\pi}{3}]$
