To illustrate computing double integrals asiterated integrals, we start withthe simplest example of a double integral over a rectangle and thenmove on to an integral over a triangle.
Example 1
Compute the integral \begin{align*} \iint_\dlr x y^2 dA\end{align*}where $\dlr$ is the rectangle defined by $0 \le x \le 2$ and $0 \le y \le 1$ pictured below.
Solution: We will compute the double integral as theiterated integral\begin{align*} \int_0^1 \left( \int_0^2 xy^2 dx\right) dy.\end{align*}We first integrate with respect to $x$ inside the parentheses.Similar to the procedure withpartial derivatives,we must treat $y$ as aconstant during this integration step. Since for any constant $c$, the integral of $cx$ is$cx^2/2$, we calculate\begin{align*} \int_0^1 \left( \int_0^2 xy^2 dx\right) dy &= \int_0^1 \left(\left. \frac{x^2}{2} y^2 \right|_{x=0}^{x=2} \right)dy\\ &= \int_0^1 \left( \frac{2^2}{2} y^2 - \frac{0^2}{2} y^2 \right)dy \\ &= \int_0^1 2 y^2 dy.\end{align*}Note that in the first line above, we wrote the limits as $x=2$ and$x=0$ so it is unambiguous that $x$ is the variable we justintegrated.
To finish, we need to compute the integral with respect to $y$, whichis simple. Since $x$ is gone, it's just a regular one-variableintegral. We calculate that our double integral is\begin{align*} \iint_\dlr x y^2 dA &= \int_0^1 2 y^2 dy\\ &= \left. \frac{2y^3}{3} \right|_0^1 \goodbreak = \frac{2(1^3)}{3}- \frac{2(0^3)}{3} \goodbreak =\frac{2}{3}.\end{align*}
To double check our answer, we can compute the integral in the other direction,integrating first with respect to $y$ and then with respect to $x$.The only trick is to remember that when integrating withrespect to $y$, we must think of $x$ as a constant. Since for any constant $c$, theintegral of $cy^2$ is $cy^3/3$, we calculate\begin{align*} \iint_\dlr x y^2 dA &= \int_0^2 \left( \int_0^1 xy^2 dy\right) dx\\ &= \int_0^2 \left(\left. x \frac{y^3}{3} \right|_{y=0}^{y=1}\right)dx\\ &= \int_0^2 \frac{x}{3} dx\\ &= \left.\left.\frac{x^2}{6} \right|_0^2\right. = \frac{4}{6} = \frac{2}{3}.\end{align*}As it must, this iterated integral gives the same answer.
Example 2
Rectangular regions are easy because the limits($a \le x \le b$ and $c \le y \le d$) are fixed, meaning the ranges of $x$ and $y$ don't depend on each other. For regions of other shapes, the range of one variable will depend on the other. Here's an examplewhere we integrate over the region defined by $0 \le x \le 2$ and $0\le y \le x/2$. The fact that the range of $y$ depends on $x$ means this region is not a rectangle. In fact, the region is the triangle pictured below.
Using the same function $f(x,y)=xy^2$ as in example 1, compute $\iint_\dlr f\,dA$where $\dlr$ is the above triangle.
Solution: A triangle is slightly more complicated than a rectangle because the limits of one variable will depend on the other variable. For the triangle defined by $0 \le x \le 2$and $0 \le y \le x/2$, the limits of $y$ depend on $x$. For a givenvalue of $x$, $y$ ranges from 0 to $x/2$, as illustrated above by thevertical dashed line from $(x,0)$ to $(x,x/2)$.
In a double integral, the outer limits must be constant, but the innerlimits can depend on the outer variable. This means, we must put $y$as the inner integration variables, as was done in the second way ofcomputing Example 1. The only difference from Example 1 is that theupper limit of $y$ is $x/2$. The double integral is\begin{align*} \iint_\dlr x y^2 dA &= \int_0^2 \left(\int_0^{x/2} xy^2 dy \right) dx\\ &=\int_0^2 \left(\left.\frac{x}{3} y^3 \right|_{y=0}^{y=x/2}\right) dx\\ &=\int_0^2 \left( \frac{x}{3} \left(\frac{x}{2}\right)^3 -\frac{x}{3} 0^3 \right) dx\\ &= \int_0^2 \frac{x^4}{24} dx\\ &= \left.\left.\frac{x^5}{5\cdot 24}\right|_0^2\right. = \frac{32}{5 \cdot 24} = \frac{4}{15}.\end{align*}
Example 2'
Now compute the integral over the same triangle $\dlr$, but make $y$be the outer integration variable.
Solution: Now we need to give constant limits for $y$. Asillustrated below, the total range of $y$ within the triangle is betweenbetween $0$ and $1$. Then, for a given value of $y$, $x$ takes onvalues between $2y$ and $2$ (as shown by the horizontal dashed linebetween $(2y,y)$ and $(2,y)$). Hence, we can describe the triangle by$0 \le y \le 1$ and $2y \le x \le 2$.
Is it confusing that the limits of $x$ are $2y \le x \le 2$ ratherthan $0 \le x \le 2$ (which would more closely parallel the aboveExample 2)? If we let $x$ range from $0$ to $2y$, then the trianglewould be the upper-left triangle in the above picture. We want tocompute the integral over the region $\dlr$, which is the lower-righttriangle shaded in red. In this triangle, $y \lt x/2$ (as used above inExample 2) which means that for this example, we must use $x > 2y$.
The double integral is similar to the first way of computing Example1, with the only difference being that the lower limit of $x$ is$2y$. The integral is\begin{align*} \iint_\dlr x y^2 dA &= \int_0^1 \left( \int_{2y}^2 xy^2 dx \right)dy\\ &= \int_0^1\left(\left.\frac{x^2y^2}{2} \right|_{x=2y}^{x=2}\right) dy\\ &= \int_0^1 \left( 2y^2 - \frac{(2y)^2 y^2}{2}\right) dy\\ &= \int_0^1 \left( 2y^2 - 2y^4\right) dy\\ &= 2 \left[ \frac{y^3}{3} - \frac{y^5}{5} \right]_0^1\\ &= 2 \left(\frac{1}{3} - \frac{1}{5} -(0-0)\right)\\ &= 2 \cdot \frac{2}{15} \goodbreak = \frac{4}{15}.\end{align*}Thankfully, this does agree with the answer we obtained in Example 2.
More examples
To go from Example 2 to Example 2', we “changed the order ofintegration.” You can see more examples of changing the order of integration in doubleintegrals. You can also see more double integral examples from the special cases of interpreting double integrals as area and double integrals as volume.
