Doubts regarding the change of order of integration $$\int_0^1\int_0^x \sqrt{x+y^2}\,dydx = \int_0^x\int_0^1 \sqrt{x+y^2}\,dxdy$$
Are these two integrals equivalent to each other? I assumed they weren't after imagining that one should also change the order of integration (in that case, analysing the region and changing the limits of integration). Am I confused about something, or does my reasoning make sense? If I'm right, I'd like to understand why you have to change the limits if you change the order of integration (I know that one can use Fubini's theorem on rectangular regions). Sorry if this isn't in the correct place, I have never posted here.
 A: We can write OPs left-hand side as
\begin{align*}
\color{blue}{\int_0^1\int_0^x \sqrt{x+y^2}\,dydx}&=\int_{0\leq x\leq 1}\int_{0\leq y\leq x}\sqrt{x+y^2}\,dydx\\
&=\int_{0\leq y\leq 1}\int_{y\leq x\leq 1}\sqrt{x+y^2}\,dxdy\\
&\,\,\color{blue}{=\int_0^1\int_{y}^1\sqrt{x+y^2}\,dxdy}\tag{1}
\end{align*}

*

*The left-hand side of (1) gives $\int_0^1\int_0^x \sqrt{x+y^2}\,dydx=\int_{0}^1 f_1(x)\,dx=C\in\mathbb{R}$.


*The right-hand side of (1) gives $\int_0^1\int_{y}^1\sqrt{x+y^2}\,dxdy=\int_0^1 f_2(y)\,dy=C\in\mathbb{R}$.
On the other hand OPs right-hand side evaluates to
\begin{align*}
 \int_0^x\int_0^1 \sqrt{x+y^2}\,dxdy=\int_0^xg_1(y)\,dy=g_2(x)\in\mathbb{R}^{\mathbb{R}}
 \end{align*}
A: The limits of a double integral can be interpreted as equations. The graphs of those equations bound a region. The double integral
$$\int_{x=0}^{x=1}\int_{y=0}^{y=x} f(x,y)\,dy\,dx $$
is over the following region.

Reversing the order of integration necessitates finding limits of the form
$$ \int_{y=a}^{y=b}\int_{x=?}^{x=?} f(x,y)\,dx\,dy $$
where $a,b$ are constants and the ?? are either constants or functions of $y$.
Referring to the graph of the region of integration, we conclude that it must be
$$ \int_{y=0}^{y=1}\int_{x=y}^{x=1} f(x,y)\,dx\,dy $$
Which is actually written
$$ \int_{0}^{1}\int_{y}^{1} f(x,y)\,dx\,dy $$
