reverse the order of integration How do you reverse the order of this integral into $dy\,dx$? I feel like you need two separate ones but I don't know how to do it:
$$\int_0^3\int_\sqrt{y}^3 f(x,y) \, dx \, dy$$
thanks
 A: If you draw the region, you will see that it looks like a trapezoid, and it can be divided into two regions $A$ and $B$. 
Over region $A$: $\displaystyle \int_{0}^\sqrt{3} \int_{0}^{x^2} f(x,y)\,dy\, dx$,
Over region $B$: $\displaystyle \int_{\sqrt{3}}^3 \int_{0}^3 f(x,y)\,dy\,dx$,
and the answer is the sum of the two integrals above.
A: This is not a complete solution, but hopefully will be enough to get you started.
Read the expression from the outside inwards:  You have $y$ ranging from 0 to 3, and for each $y$ value in that interval, $x$ ranges from $\sqrt{y}$ to 3.  So try graphing the curve $x=\sqrt{y}$ and try to shade in the region this describes.
Once you have the region sketched, try to describe it again, but in the other order:  first in terms of $x$ (what is the lowest value of $x$ in the region?  What is the highest value of $x$ in the region), and then, for each $x$, describe how $y$ varies (keeping in mind that the upper or lower limits for $y$ might be expressed in terms of $x$).
Hope that helps!
