Order of Integration

Some regions can be viewed as both Type I or Type II.  In that case we can set up an iterated integral in two ways.  Depending on the integrand, one can be a lot easier than the other.

Sometimes you're given an impossible-looking iterated integral, and you can solve it by changing the order of integration.  Here are the steps to change the order of integration.

  1. Begin with an iterated integral that is a double integral over some region $D$.
  2. If you are not given $D$, then determine $D$ from the limits of integration.
  3. Express $D$ as the other type of region (Type II if it was originally set up as Type I, and Type I if it was originally set up as Type II).
  4. Rewrite the integral over $D$ as an interated integral in the other way.  (If had originally been $dx \,dy$, it should now be $dy \,dx$, and vice-versa.)
  5. Evaluate the new iterated integral.

Some examples are worked in detail in this video.


Example 1: Evaluate the integral $$I \ = \iint_D\, x\,\sqrt{1+y^3}\, dA$$ when $D$ is the triangular region shown here, enclosed by the $y$-axis and the lines $$y \ = \ \frac{1}{3}x\,, \qquad y \ = \ 2\,.$$


Bad Solution 1:  Fix $x$ and integrate with respect $y$ along the vertical red line. Then $\displaystyle I \ = \ \int_0^6 \left(\int_{x/3}^2\, x\,\sqrt{1+y^3}\, dy\right)\,dx\,.$  The trouble is that the inner integral involves requires evaluating the integral $\displaystyle\int_{x/3}^2\, \sqrt{1+y^3}\, dy\,.$  Nothing you've learned so far in calculus will work here, so we change the order of integration to see if that helps.


Good Solution 1:  Fix $y$ and integrate with respect to $x$ along the black line.  Then $\displaystyle I \ = \ \int_0^2 \left(\int_{0}^{3y}\, x\,\sqrt{1+y^3}\, dx\right)dy\,.$  The inner integral has $\sqrt{1+y^3}$ as a constant, and involves evaluating the integral $$\displaystyle\int_{0}^{3y}\, x\sqrt{1+y^3}\, dx\ = \ \Bigl[\,\frac{1}{2}x^2\sqrt{1+y^3}\, \Bigr]_0^{3y} \ = \ \frac{9}{2}y^2\sqrt{1+y^3}\,.$$  In this case, $$I \ = \ \frac{9}{2} \int_0^2\, y^2 \, \sqrt{1+y^3}\,dy\ \overset{\fbox{$ \,\,u\,=\,1+y^3,\\ du\,=\,3y^2\,dy\\u(0)=1\\u(2)=9$}\\}{=} \frac{3}{2}\int_1^9 u^{1/2}\,du= \frac{3}{2}\frac{2}{3}u^{3/2}\Bigr|_1^{9}=26\,.$$


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Reversing the order of integration in a double integral requires creating a graph of the region of integration.  Then it's a matter of algebra and inverse functions.


Example 2:  Reverse the order of integration in the iterated integral $$I \ = \ \int _0^2\left(\int_{x^2}^4\, f(x,\,y)\, dy\right)dx\,.$$

DO:  Graph the region of integration, which we'll call $D$.


Solution 2:  The region of integration is the set $$D \ = \ \Bigl\{\,(x,\,y) : \, x^2 \le y \le 4\,, \ \ 0 \le x \le 2\,\Bigr\}$$ which we determine from the limits of integration of $I$.  The graph of $D$ is given to the right.  The given iterated integral fixes $x$ and integrates with respect to $y$ along the vertical black line.  To reverse the order of integration we need to fix $y$ and integrate with respect to $x$ along the red line.

DO:  Set up the new iterated integral before proceeding.


The horizontal red line in graph of $D$ shows the lower limit of integration is $x = 0$.  The red line ends at the parabola $y = x^2$, which can be written as a function of $y$ as $x = \sqrt{y}$; this tells us the upper limit of integration $ x =\sqrt{y}$.  We also note that $y$ goes from $y=0$ to $y=4$.  Thus $D$ can also be written as $$D \ = \ \Bigl\{\,(x,\,y) : \, 0 \le x \le \sqrt{\,y }, \ \ 0 \le y \le 4\,\Bigr\}\,.$$ Rewriting $I$ after changing the order of integration, we get $$I \ = \ \int _0^4\left(\int_0^{\sqrt{\,y}}\, f(x,\,y)\, dx\right)\,dy\,.$$

Warning: The tricky part of swapping the order of integration is rewriting the limits of integration. This involves studying the region of integration. $$\int_a^b \int_{y=g(x)}^{h(x)}\, f(x,y) \,dy\, dx$$ does NOT become $$\int_{g(x)}^{h(x)} \,\int_a^b f(x,y) \,dx\, dy \hbox{ !!}$$