In calculus, the term ‘order of integration’ pertains to the topic of multiple integrals. When solving multiple integrals, (say double or triple integrals), we evaluate for variable ‘x’ (say) first and then for variable ‘y’, depending on how the question is given to us.

However it is possible to change (in case of triple integrals) or reverse (in case of double integrals) the order of integration. Meaning, we can evaluate using variable y first and then using variable x in case of double integrals. The limits of the integral would change accordingly.

Statement for the order of integration rule can be stated symbolically as follows:
$\int_{a}^{b}\int_{f_1(y)}^{f_2(y)}f(x,y)dx dy$

=$\int_{c}^{d}\int_{g_1(y)}^{g_2(y)}f(x,y)dy dx$
Note that in the first expression, the limits of the dx are functions of y,where as in the second expression the limits of dx are constants c and d. Similarly in the first expression the limits of dy are constants and in the second expression they are functions of x. Similar symbolic representation of changing order of integration triple integrals is also possible. 

The order of integration rule and the steps for changing the order of integration may vary slightly for double and triple integrals. Let us take a look at them one by one.
Many a times while solving double integrals, we encounter situations, where in it seems difficult or impossible to iterate in one direction. In such situations if we reverse the order of integration, then the integration becomes very easy.  

This can be best understood with the help of an example:

Example 1: Consider the following double integral:
I = $\int_{1}^{2}\int_{0}^{lnx} f(x,y)dy dx$

Here, the outer limits belong to variable x. Thus 1 $\leqslant$ x $\leqslant$ 2. The other  limit is for the variable y. So, 0 = y = ln(x).

It is important to note here, that the limits of x are constants, where as those of y are functions of x. We saw a similar situation in our symbolic representation earlier as well. We will now use all this information to graph the region that we are trying to integrate.

It would look as follows:
Order of Integration Graph
The said region is the pink region R. If we now wish to reverse the order, it would look as follows:
I = $\int_{0}^{lnx} \int_{1}^{2}$ f(x,y)dy dx$
So you see that the dy dx of the end has not become dx dy. However we are still to find the new limits of integration. So now this time the outer limits would belong to y and so we need constants for that now. From the above graph we can see that the smallest value of y is 0 where as the largest value of y is ln(2). Thus the limits of y would be from 0 to ln(2). (We can see that from the graph).

Now, coming to the inner limits since we already established that the outer limits belong to y, so obviously the inner limits would belong to x.

Also, since these are inner limits, we need x in terms of some function of y. The function given to us in the question is y = ln(x) (the upper limit of y given in the original question). Solving that for x we have, x = $e^y$. From the graph we also see that the right hand side limit of the x axis for the region R is the line x=2. Thus the upper limit of x would be 2 and the lower limit would be $e^y$.

Therefore now our new integral with reversed order would looks follows:

I = $\int_{0}^{ln2}\int_{e^y}^{2} f(x,y)dy dx$

Thus we see that we have successfully changed the order of integration for the given example problem.
We are quite familiar with the integrate over a two-dimensional region. Now let us move on to integrating over a three-dimensional region. In 2D we used double integral to integrate a function and three integrals to integrate function in 3D.
The method of changing order of integration in triple integrals remains the same as in two dimension. We just need to change the order in pairs. Thus to change the order of all three, we need to do the above process twice, in two pairs.Example: Write the integral $\int_0^2$$\int_y^2$$\int_{y^2+1}^5$ dzdxdy in 2 other ways.

Solution: We can see that inner two integrals have bounds only depending on y so we can easily switch them.

 $\int_0^2$ $\int_{y^2+1}^5$ $\int_y^2$ dx dz dy

Now let us try to write integral as $\int$$\int$$\int$ dz dy dx
Let x varies from 0 to 2 and let y varies from 0 to x

From the limits we come to know that the x bounds, the plane x = y and x = 2. then let z part is lies between y$^2$ + 1 and 5

So integral can be written as

$\int_0^2$$\int_0^x$$\int_{y^2+1}^5$ dz dy dx