Theorem for a $2$-dimensional (easy) integral with variable boundary I have a problem with the integral $$\int_0^1 \int_0^{t^2}x^2 t^2  dx dt .$$The problem I have with this integral is that in the description of the exercise it said that we should evaluate this (I did that) and after that we should change the order of integration - it also said that we should beware of the limits. 
MY question is now: What exactly should I change when integrating ? Should I integrate
i) $\int_0^1 \int_0^{t^2}x^2 t^2 dt dx $ or
ii)  $ \int_0^{t^2} \int_0^1 x^2 t^2 dt dx $ or 
iii)  $ \int_0^{t^2} \int_0^1 x^2 t^2 dx dt $ 
(all possible combinations of changing the integral and the $dx,dt$'s) ?
And how does this relate to my first integral ? 
We did a theorem about changing the order of integration, but only if the domain of integration is fixed (whereas here it depends on $t$) - if I could apply that theorem, then iii) would be the correct integral that is equal to the first one.
Please tell me what the scope of this exercise was, since our professor didn't give us any explanation.
 A: When you "change the order of the variables of integration", you are calculating an integral in a slightly different way. 
When we write $\int_a^b \int_c^d dt dx$ one first imagines the $x$ running from $a$ to $b$, and for each value of $x$, you have to let $t$ run from $c$ to $d$ (note that here $c$ and $d$ may depend on $x$, so they maybe functions $c(x)$, and $d(x)$).  
On the other hand, when we write $\int_e^f \int_g^h dx dt$ one first imagines the $t$ running from $e$ to $f$, and for each value of $t$, you have to let $t$ run from $g$ to $h$ (note that here $g$ and $h$ may depend on $t$, so they maybe functions $g(t)$, and $h(t)$).
In your case, the bounds in the integral $\int_0^1 \int_0^{t^2} (\ \cdots\ ) dx dt$ determine a region in the $t$,$x$ plane (here I am going to choose to put $t$ in the horizontal axis and $x$ in the vertical axis). The bounds for $t$ say "run $t$ between $0$ and $1$" and the bounds for $x$ say, "for each value of $t$, run $x$ between $x=0$ and $x=t^2$". If you follow this recipe, you should see (draw it!) that the region is the area under the parabola $x=t^2$, between $t=0$ and $t=1$.
Now, if you want to change the order of the variable of integration, you need to change your point of view and your description of the area that you are integrating over. This time, we are going to run over values of $x$ in the vertical axis, and then determine what values of $t$ we should consider. If you have a good drawing of the area, you shall see that $x$ should run from $0$ to $1$, and for each value of $x$, the values of $t$ run from $t=\sqrt{x}$ to $t=1$. This translates into the integral
$$\int_0^1 \int_{\sqrt{x}}^1 (\ \cdots \ ) dt dx.$$    
A: In the integral $$\int_0^1\int _0^{t^2} x^2 t^2 dx  {dt}  ,$$ the region of integration is the gray region shown below:

In the above integral, we think of the region as being generated by   vertical line segments (a typical one is the blue line above) that range  from $t=0$ to $t=1$.
The inner integral above corresponds to integrating over a typical vertical line segment. It is a  "$dx$ integral", since the line segment  is vertical. The limits on the inner integral are from where a typical blue line segment starts to where it ends. So you have
$$ \int_{x=0}^{x=t^2} x^2 t^2  dx  $$
Then you integrate with respect to the $t$ variable, which corresponds to "summing the vertical line segment integrals" as they range from $t=0$ to $t=1$.
 $$ \int_{ 0}^{1}{t^8\over3}  dt   .$$

If you want to switch the order of integration, you need to think of the region as being generated by horizontal line segments (a typical one is the green line above)  that range from $x=0$ to $x=1$. 
You first integrate over a horizontal line segment (so it's a $dt$ integration  now and the limits are from where a typical line starts to where it ends):
$$
\int_{t=\sqrt{x}}^{t=1} t^2x^2   dt.
$$
Then integrate the "horizontal line integrals" from $x=0$ (where they start) to $x=1$ (where they end).
$$
\int_0^1\int_{ \sqrt{x}}^1 t^2x^2   dt dx
$$ 
A: The two answers already here are great. I just wanted to say that with some interpretation you can still apply your theorem, which only mentioned constant bounds of integration. Your integral (assuming you've accidentally switched $dt$ and $dx$ in your statement) is equal to
\[
\int_0^1\int_0^1 x^2t^2f(x, t)\, dxdt
\]
where
\[
f(x, t) = \begin{cases}
1 & x \leq t^2 \\
0 & \text{else}
\end{cases}.
\]
This is the characteristic function of the region of interest in the plane.
A: First, when you change the order of integration, you wind up integrating over $t$, then $x$.  So (as a matter of formal notation, though it really doesn't matter), you should have $dx dt$ inside the integral before the change and $dt dx$ after.  
Second, you need to consider exactly what region you are integrating over.  The region is the part of the $x-t$ plane that lies between the parabola defined by $x=t^2$, the $t$-axis, the $x$-axis, and the line $t=1$.  So you need your new integral to match those bounds.  You need to define, for each value of $x$, the permitted values of $t$.  $x$ itself is going to range from $0$ to $1$.  Can you see why?
For each value of $x$, you know that $x<=t^2$.  You also know that $t^2 <= 1$.  So $t$ lies in the range $[ \sqrt{x}, 1]$ and your integral looks like

$ \int_0^1 \int_{\sqrt{x}}^1 x^2 t^2 dt dx $.

A: The expression
$$ \int_0^{t^2} \int_0^1 x^2 t^2 \;dt \;dx $$
is meaningless.  The variable $t$ goes from something to something in the inside integral.  If you write
$$
\int_0^1 \int_0^{t^2} x^2 t^2 \;dt\; dx,
$$
that is also meaningless, because you're saying $t$ is something that goes from $0$ to $t^2$. If you write
$$
\int_0^1 \int_0^{t^2} \cdots\cdots\;dx \;dt
$$
then that means something.  The point is that firstly, $t$ goes from $0$ to $1$, and then for any particular value of $t$, $x$ goes from $0$ to $t^2$.
Google the terms "free variable" and "bound variable".
