Often, one encounters a double integral of the form: $$\int_{x=a}^b \int_{y=\alpha(x)}^{\beta(x)} f(x,y)\, dy\, dx,$$ with $\alpha,\beta$ functions from $\mathbb{R}$ to $\mathbb{R}$ and $f(x,y)$ a function s.t. Fubini's theorem may be applied.
In my case, these functions $\alpha$ and $\beta$ have mostly been linear function or simple polynomials. In this case, I just draw the area on which I need to integrate and reason what the new boundaries should be after applying Fubini.
I was now wondering if there is some general result, like If $\alpha$ and $\beta$ satisfy condition $A$ then the integrals can be swapped and the integrals become $\int_{y=a}^b \int_{\gamma(y)}^{\delta(y)} f(x,y) dx dy$ with $\gamma$ and $\delta$ given by ...
Let $B\subset{\mathbb R}^2$ be a reasonable domain, let $B':=\{x\,|\;\exists (x,y)\in B\}$ be the projection of $B$ onto the $x$-axis, and let $B_x:=\{y\,|\,(x,y)\in B\}$ be the set of points cut out of $B$ by the ordinate erected at $x$. Then Fubini's theorem says that $$\int_B f(x,y)\>{\rm d}(x,y)=\int_{B'}\int_{B_x} f(x,y) dy\>dx\ .$$ By symmetry, one also has the formula $$\int_B f(x,y)\>{\rm d}(x,y)=\int_{B''}\int_{B_y} f(x,y) dx\>dy\ ,$$ where $B''$ and $B_y$ are defined analogously.
The above is universally true, and valid even for "complicated" domains $B$. But if, e.g., $B$ has "holes", then the $B_x$ (resp. $B_y$) will not be intervals $[\alpha(x),\beta(x)]$ as in the case of a convex $B$. Instead $B_x$ might be a union of intervals, depending on $x$, so that one has to partition the computation. In cases where the $B_y$ are intervals it may be more advantageous to resort to the second formula.
For the determination of the $B_x$ (resp., the $B_y$) you don't need general theorems, but algebraic skills and some expertise. Always draw a figure!