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Order of IntegrationSome 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 impossiblelooking iterated integral,
and you can solve it by changing
the order of integration. Here are the steps
to change the order of integration.
Some examples are worked in detail in this video.
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, $$I \ = \ \frac{9}{2} \int_0^2\, y^2 \, \sqrt{1+y^3}\,dy\ \overset{\fbox{$ \,\,u\,=\,1+y^3,\\ du\,=\,3y^2\,dy\\u(0)=1\\u(2)=9$}\\}{=} \frac{3}{2}\int_1^9 u^{1/2}\,du= \frac{3}{2}\frac{2}{3}u^{3/2}\Bigr_1^{9}=26\,.$$  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 $D$.
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 $$I \ = \ \int _0^4\left(\int_0^{\sqrt{\,y}}\,
f(x,\,y)\, dx\right)\,dy\,.$$
