Are these boundaries true? I want to find the double integral 
$$\iint xy \, dx\,dy $$ 
over the region
bounded by the positive $y$- axis, the line $y=\sqrt3 \ x$ and the circle
 $x^2+y^2=4$.
My solution is that  $x=0$ to $x=(1/\sqrt3) y$, and $y=0$ to $y=2$.
Is that correct?
 A: By switching to polar coordinates, you integral is given by:
$$ \int_{\pi/3}^{\pi/2}\int_{0}^{2}\rho^3\sin(\theta)\cos(\theta)\,d\rho\,d\theta=2\int_{\pi/3}^{\pi/2}\sin(2\theta)\,d\theta=\cos(2\pi/3)-\cos(\pi)=\frac{1}{2}.$$
A: 
My solution is that  $x=0$ to $x=(1/\sqrt3) y$, and $y=0$ to $y=2$.

If you express it in that kind of language, it would be best to be explicit that you're putting the integral with respect to $y$ on the outside.
This would be
$$
\int_0^2 \left( \int_0^{y/\sqrt 3} xy \, dx \right) \, dy
$$
and that is not correct.  It ignores the equation $x^2+y^2=4$.
If $x^2+y^2 = 4$ and $x\sqrt3 = y$, then $3x^2 + x^2 = 4$ then $x= \pm 1$.  Since you say the "positive" $y$-axis, we need $x=+1$.
The equation $y^2+x^2=4$, is equivalent to $y = \pm\sqrt{4-x^2}$, and conjoining that with $y\ge 0$ we have $y = \sqrt{4-x^2}$.
So we can say $x$ goes from $0$ up to $1/\sqrt3$, and then for any fixed value of $x$, the other variable $y$ goes from $x\sqrt3$ up to $\sqrt{4-x^2}$, so we have
$$
\int_0^1 \left( \int_{x\sqrt3}^{\sqrt{4-x^2}} xy\, dy \right) \, dx
$$
(If there is any question about which of the two bounds on the inside integral is bigger, draw a picture.) Details below.
If you want to do it in the other order, then it's more complicated.  You'd have
$$
\int_0^1 \left( \int_0^{y/\sqrt3} xy \, dx \right) \, dy + \int_1^2 \left( \int_0^{\sqrt{4-y^2}} xy \, dx \right) \, dy.
$$
In order to actually evaluate the integral, I would not do it like this; I would use polar coordinates.
Details: We have $\displaystyle \int_0^1 \left( \int_{x\sqrt3}^{\sqrt{4-x^2}} xy\, dy \right) \, dx$.  The inside integral is $$\int_{x\sqrt3}^{\sqrt{4-x^2}} xy \, dy = \left[ \frac{xy^2} 2 \right]_{y:=x\sqrt3}^{y:=\sqrt{4-x^2}}$$ $$= \frac{x(4-x^2)} 2 - \frac{x\cdot 3x^2} 2 = 2x - 2x^3.$$ Then the outside integral becomes
$$ \int_0^1 \left( 2x - 2x^3 \right) \, dx = \left[ x^2 - \frac{x^4} 2 \right]_0^1 = \frac 1 2$$
But (as I said above) this is not how I would normally do this.  I would use polar coordinates.
