integration over an area using Fubini 
I want to calculate the following areas of f over S :  
a) $S$ is given by the area enclosed by : $y=1, x=3, y=2x$ and  $\displaystyle f(x,y) = \frac{x^{2}}{y^{2}}$.
  b) $S$ is given by the area enclosed by : $x=y^{2}-4$, $x=3y$ and $f(x,y)= 1$

I calculated the following for a)$$\int_{\frac{1}{2}}^{3}\left(\int_{1}^{2x}\frac{x^{2}}{y^{2}} dy\right) dx.$$
for b) I switched $x$ with $y$ and calculated this:  $$\int_{-1}^{4}\left(\int_{-3}^{x^{2}-4} 1dy\right)dx.$$
Is this correct? I draw $S$ and look at the intersection points. Still I have problem determining the right boundaries for my integrals. Can you give me a hint/idea what is a good way to find boundaries?
Merci. 
 A: Part a) looks good.
For part b), just draw the given region (without switching $x$ and $y$).
The region in question  is bounded by a parabola opening to the right with vertex at $(-4,0)$ and the line passing through the origin of slope $1/3$. The points of intersection are $(-3,-1)$ and $(12,4)$. So, the region is bounded on the left by the parabola and bounded on the right by the line.
You would run into problems integrating with respect to $y$ first, because the functions giving the  top and bottom bounds of the "representative vertical line segment at $x$" changes across $x=-4$ to $x=12$.
Things are rosy if you integrate with respect to $x$ first. Here, the inner integral will have limits that correspond to the left and right endpoints of a representative horizontal line segment (the maroon line segment in the diagram  below):
$$
\int_{x=y^2-4}^{x=3y} 1\,dx.
$$ 
You then integrate the "horizontal line segment integrals" as they range from $y=-1$ to $y=4$:
$$
\int_{-1}^4 \int_{ y^2-4}^{3y} \,dx \,dy.
$$


