Use Fubini's theorem to reverse the order of the double integral 
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*Use Fubini's theorem to reverse the order of the double integral (the reversed integral may split into a sum of multiple pieces):

a) $\int_0 ^4 \int_0 ^{\sqrt{x}} f(x,y) dy dx$
b)  $\int_0 ^2 \int_x ^{3} f(x,y) dy dx$
c) $\int_{-1} ^2 \int_0 ^{1-y^2} f(x,y) dx dy$
In part $c)$ be careful about signs!
My attempt
a) 
Boundary of number at $x$: $0\leq x\leq 4$
Boundary of function at $y$: $0\leq y\leq \sqrt{x}.$
Now consider,
boundary of number at $y$: $0\leq y\leq 2$
boundary of function at $x$: $y^2\leq x\leq 4.$
So by the Fubini's
$\int_0 ^4 \int_0 ^{\sqrt{x}} f(x,y) dy dx=\int_0 ^2 \int_{y^2}^4 f(x,y) dx dy.$
b)
Boundary of number at $x$: $0\leq x\leq 2$,
boundary of function at $y$: $x\leq y\leq 3.$
Now, consider
boundary of number at $y$: $0\leq y\leq 3$
boundary of function at $x$: $0\leq x\leq y$
So, by the Fubini's theorem:
$\int_0 ^2 \int_x ^{3} f(x,y) dy dx=\int_0 ^3 \int_0 ^{y} f(x,y) dy dx.$
c) I couldn't draw graph, may you draw and add here?
May you check $a), b)$ and may you help for $c)$?

The graph and working for (a) are correct. The graph for (b) is wrong and should look like

The reversed integral is $\int_0^3\int_0^{\min(2,y)}f(x,y)\,dx\,dy$.
For (c) the graph looks like this:

where the negative sign indicates a reversed orientation. This integral needs to be split when reversed, and the result is
$$\int_0^1\int_{-\sqrt{1-x}}^{\sqrt{1-x}}f(x,y)\,dy\,dx-\int_{-3}^0\int_{\sqrt{1-x}}^2f(x,y)\,dy\,dx$$
