As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.
Reverse the order of integration in the iterated integral [latex]\displaystyle\int_{x=0}^{x=\sqrt2}\displaystyle\int_{y=0}^{y=2-x^2}xe^{x^2}dydx[/latex]. Then evaluate the new iterated integral.
The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to Figure 10.
Figure 10. Converting a region from Type I to Type II.
We can see from the limits of integration that the region is bounded above by [latex]y=2-x^{2}[/latex] and below by [latex]y=0[/latex], where [latex]x[/latex] is in the interval [latex][0,{\sqrt{2}}][/latex]. By reversing the order, we have the region bounded on the left by [latex]x=0[/latex] and on the right by [latex]{x} = {{\sqrt{2-y}}}[/latex] where [latex]y[/latex] is in the interval [latex][0,2][/latex]. We solved [latex]y=2-x^2[/latex] in terms of [latex]x[/latex] to obtain [latex]{x} = {{\sqrt{2-y}}}[/latex].
Hence
[latex]\begin{align} \displaystyle\int_0^{\sqrt2}\displaystyle\int_0^{2-x^2}xe^{x^2}dydx&=\displaystyle\int_0^2\displaystyle\int_0^{\sqrt{2-y}}xe^{x^2}dxdy \,\,\,\,\,\text{ Reverse the order of integration then use substitution} \\ &=\displaystyle\int_0^2\left[\frac12ex^2\bigg|_0^{\sqrt{2-y}}\right]dy=\displaystyle\int_0^2\frac12(e^{2-y}-1)dy=-\frac12(e^{2-y}+y)\bigg|_0^2 \\ &=\frac12(e^2-3). \end{align}[/latex].
Consider the iterated integral [latex]\underset{R}{\displaystyle\iint}{f(x,y)}{dx} \ {dy}[/latex] where [latex]{z} = {f(x,y)} = {x-2y}[/latex] over a triangular region [latex]R[/latex] that has sides on [latex]x=0,\text{ }y=0[/latex] and the line [latex]x+y=1[/latex]. Sketch the region, and then evaluate the iterated integral by
a. integrating first with respect to [latex]y[/latex], and then,
b. integrating first with respect to [latex]x[/latex].
A sketch of the region appears in Figure 11.
Figure 11. A triangular region [latex]R[/latex] for integrating in two ways.
We can complete this integration in two different ways.
a. One way to look at it is by first integrating [latex]y[/latex] from [latex]y=0[/latex] to [latex]y=1-x[/latex] vertically and then integrating [latex]x[/latex] from [latex]x=0[/latex] to [latex]x=1[/latex]:
b. The other way to do this problem is by first integrating [latex]x[/latex] from [latex]x=0[/latex] to [latex]x=1-y[/latex] horizontally and then integrating [latex]y[/latex] from [latex]y=0[/latex] to [latex]y=1[/latex]:
