So far, we have seen how to set up a double integral and how to obtain an approximate value for it. We can also imagine that evaluating double integrals by using the definition can be a very lengthy process if we choose larger values for [latex]m[/latex] and [latex]n[/latex]. Therefore, we need a practical and convenient technique for computing double integrals. In other words, we need to learn how to compute double integrals without employing the definition that uses limits and double sums.
The basic idea is that the evaluation becomes easier if we can break a double integral into single integrals by integrating first with respect to one variable and then with respect to the other. The key tool we need is called an iterated integral.
iterated integral
Assume [latex]a[/latex], [latex]b[/latex], [latex]c[/latex], and [latex]d[/latex] are real numbers. We define an iterated integral for a function [latex]f(x, y)[/latex] over the rectangular region [latex]R = [a,b] \times [c,d][/latex] as:
The notation [latex]\displaystyle\int_a^b\left[\displaystyle\int_c^df(x,y)dy\right] dx[/latex] means that we integrate [latex]f(x,y)[/latex] with respect to [latex]y[/latex] while holding [latex]x[/latex] constant. Similarly, the notation [latex]\displaystyle\int_c^d\left[\displaystyle\int_a^bf(x,y)dx\right] dy[/latex] means that we integrate [latex]f(x,y)[/latex] with respect to [latex]x[/latex] while holding [latex]y[/latex] constant.
The fact that double integrals can be split into iterated integrals is expressed in Fubini’s theorem. Think of this theorem as an essential tool for evaluating double integrals.
Fubini’s theorem
Suppose that [latex]f(x,y)[/latex] is a function of two variables that is continuous over a rectangular region [latex]R = { (x,y)\in \mathbb{R}^2 \mid a \leq x \leq b,c \leq y \leq d }[/latex]. Then the double integral of [latex]f[/latex] over the region equals an iterated integral,
More generally, Fubini’s theorem is true if [latex]f[/latex] is bounded on [latex]R[/latex] and [latex]f[/latex] is discontinuous only on a finite number of continuous curves. In other words, [latex]f[/latex] has to be integrable over [latex]R[/latex].
Figure 1 illustrates both orders of integration visually, showing how we can integrate in either order to find the volume.
Figure 1. (a) Integrating first with respect to [latex]y[/latex] and then with respect to [latex]x[/latex] to find the area [latex]A(x)[/latex] and then the volume [latex]V[/latex]; (b) integrating first with respect to [latex]x[/latex] and then with respect to [latex]y[/latex] to find the area [latex]A(y)[/latex] and then the volume [latex]V[/latex].Use Fubini’s theorem to compute the double integral [latex]\underset{R}{\displaystyle\iint}{f(x,y)dA}[/latex] where [latex]f(x,y) = x[/latex] and [latex]R = [0,2] \times [0,1][/latex].
Fubini’s theorem offers an easier way to evaluate the double integral by the use of an iterated integral. Note how the boundary values of the region [latex]R[/latex] become the upper and lower limits of integration.[latex]\hspace{5cm}\begin{align} \underset{R}{\displaystyle\iint}{f(x,y)dA}&=\underset{R}{\displaystyle\iint}{f(x,y)dxdy} \\ &=\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=0}^{x=2}xdxdy \\ &=\displaystyle\int_{y=0}^{y=1}\left[\frac{x^2}{2}\Bigg|_{x=0}^{x=2}\right]dy \\ &=\displaystyle\int_{y=0}^{y=1}2dy=2y\big|_{y=0}^{y=1}=2. \end{align}[/latex]
The double integration in this example is simple enough to use Fubini’s theorem directly, allowing us to convert a double integral into an iterated integral. Consequently, we are now ready to convert all double integrals to iterated integrals and demonstrate how the properties listed earlier can help us evaluate double integrals when the function [latex]f(x, y)[/latex] is more complex. Note that the order of integration can be changed.
Evaluate the double integral [latex]\underset{R}{\displaystyle\iint}{(xy-3xy^2)dA}[/latex] where [latex]{R} = {\left \{ (x,y) \mid 0 \leq x \leq 2,1 \leq y \leq 2 \right \}}[/latex].
This function has two pieces: one piece is [latex]xy[/latex] and the other is [latex]3xy^{2}[/latex] Also, the second piece has a constant[latex]3[/latex]. Notice how we use properties i and ii to help evaluate the double integral.[latex]\begin{align} &\,\,\,\,\,\,\,\,\underset{R}{\displaystyle\iint}{(xy-3xy^2)dA} \\ &=\underset{R}{\displaystyle\iint}{(xy)dA+\underset{R}{\displaystyle\iint}{(-3y^2)dA}} &\quad &\text{Property i: Integral of a sum is the sum of the integrals.} \\ &=\displaystyle\int_{y=1}^{y=2}\displaystyle\int_{x=0}^{x=2}xydxdy-\displaystyle\int_{y=1}^{y=2}\displaystyle\int_{x=0}^{x=2}3xy^2dxdy &\quad &\text{Convert double integrals to iterated integrals.} \\ &=\displaystyle\int_{y=1}^{y=2}(\frac{x^2}{2}y)\Bigg|_{x=0}^{x=2}dy-3\displaystyle\int_{y=1}^{y=2}(\frac{x^2}{2}y^2)\Bigg|_{x=0}^{x=2}dy &\quad &\text{Integrat with respect to }x,\text{ holding }y\text{ constant}. \\ &=\displaystyle\int_{y=1}^{y=2}2ydy-\displaystyle\int_{y=1}^{y=2}6y^2dy &\quad &\text{Property ii: Placing the constant before the integral.} \\ &=2\displaystyle\int_1^2ydy-6\displaystyle\int_1^2y^2dy &\quad &\text{Integrate with respect to }y. \\ &=2\frac{y^2}{2}\bigg|_1^2-6\frac{y^3}{3}\bigg|_1^2 \\ &=y^2\bigg|_1^2-2y^3\bigg|_1^2 \\ &=(4-1)-2(8-1) \\ &=3-2(7)=3-14=-11 \end{align}[/latex]
Over the region [latex]{R} = {\left \{ (x,y) \mid 1 \leq x \leq 3,1 \leq y \leq 2 \right \}}[/latex], we have [latex]2 \leq x^2 \leq y^2 \leq 13[/latex]. Find a lower and upper bound for the integral [latex]\underset{R}{\displaystyle\iint}{(x^2+y^2)dA}[/latex].
For a lower bound, integrate the constant function 2 over the region [latex]R[/latex]. For an upper bound, integrate the constant function 13 over the region [latex]R[/latex].
Hence, we obtain [latex]4 \leq \underset{R}{\displaystyle\iint}{(x^2+y^2)dA} \leq 26[/latex].
Evaluate the integral [latex]\underset{R}{\displaystyle\iint}{e^y\cos xdA}[/latex] over the region [latex]{R} = {\left \{ (x,y) \mid 0 \leq x \leq {\dfrac{\pi}{2},0} \leq y \leq 1 \right \}}[/latex].
This is a great example for property vi because the function [latex]f(x,y)[/latex] is clearly the product of two single-variable functions [latex]e^y[/latex] and [latex]cos \ x[/latex]. Thus we can split the integral into two parts and then integrate each one as a single-variable integration problem.[latex]\hspace{3cm}\begin{align} \underset{R}{\displaystyle\iint}{e^y\cos xdA}&=\displaystyle\int_0^1\displaystyle\int_0^{\pi/2}e^y\cos x \ dx \ dy \\ &=\left(\displaystyle\int_0^1e^ydy\right)\left(\displaystyle\int_0^{\pi/2}\cos xdx\right) \\ &=\left(e^y\bigg|_0^1\right)\left(\sin x\bigg|_0^{\pi/2}\right) \\ &=e-1 \end{align}[/latex]
As we mentioned before, when we are using rectangular coordinates, the double integral over a region [latex]R[/latex] denoted by [latex]\underset{R}{\displaystyle\iint}f(x,y)dA[/latex] can be written as [latex]\underset{R}{\displaystyle\iint}f(x,y)dxdy[/latex] or [latex]\underset{R}{\displaystyle\iint}f(x,y)dydx[/latex]. The next example shows that the results are the same regardless of which order of integration we choose.
Let’s return to the function [latex]f(x,y) = 3x^2 - y[/latex] from Example “Setting up a Double Integral and Approximating It by Double Sums” this time over the rectangular region [latex]R = [0,2] \times [0,3][/latex]. Use Fubini’s theorem to evaluate [latex]\underset{R}{\displaystyle\iint}{f(x,y)dA}[/latex] in two different ways:
First integrate with respect to [latex]y[/latex] and then with respect to [latex]x[/latex].
First integrate with respect to [latex]x[/latex] and then with respect to [latex]y[/latex].
Figure 1 shows how the calculation works in two different ways.
a. First integrate with respect to [latex]y[/latex] and then integrate with respect to [latex]x[/latex]:
b. First integrate with respect to [latex]x[/latex] and then integrate with respect to [latex]y[/latex]:
[latex]\hspace{2cm}\begin{align} \underset{R}{\displaystyle\iint}f(x,y)dA&=\displaystyle\int_{y=0}^{y=3}\displaystyle\int_{x=0}^{x=2}(3x^2-y)dx \ dy \\ &=\displaystyle\int_{y=0}^{y=3}\left(\displaystyle\int_{x=0}^{x=2}(3x^2-y)dx\right)dy=\displaystyle\int_{y=0}^{y=3}\left[x^3-xy\Bigg|_{x=0}^{x=2}\right]dy \\ &=\displaystyle\int_{y=0}^{y=3}(8-2y)dy=8y-y^2\bigg|_{x=0}^{x=2}=15. \end{align}[/latex]
Analysis
With either order of integration, the double integral gives us an answer of 15. We might wish to interpret this answer as a volume in cubic units of the solid [latex]S[/latex] below the function [latex]f(x,y)=3x^2-y[/latex] over the region [latex]R = [0,2] \times [0,3][/latex]. However, remember that the interpretation of a double integral as a (non-signed) volume works only when the integrand [latex]f[/latex] is a nonnegative function over the base region [latex]R[/latex].
In the next example we see that it can actually be beneficial to switch the order of integration to make the computation easier. We will come back to this idea several times in this chapter.
Consider the double integral [latex]\underset{R}{\displaystyle\iint}x\sin(xy)dA[/latex] over the region [latex]{R} = {\left \{ (x,y) \mid 0 \leq x \leq 3,0 \leq y \leq 2 \right \}}[/latex] (Figure 2).
Express the double integral in two different ways.
Analyze whether evaluating the double integral in one way is easier than the other and why.
Evaluate the integral.
Figure 2. The function [latex]\small{z=f(x,y)=x\sin(xy)}[/latex] over the rectangular region [latex]\small{R=[0,\pi ]\times[1,2]}[/latex].
a. We can express [latex]\underset{R}{\displaystyle\iint}x\sin(xy)dA[/latex] in the following two ways: first by integrating with respect to [latex]y[/latex] and then with respect to [latex]x[/latex]; second by integrating with respect to [latex]x[/latex] and then with respect to [latex]y[/latex].
[latex]\hspace{3cm}\begin{align} &\,\,\,\,\,\,\underset{R}{\displaystyle\iint}x\sin(xy)dA \\ &=\displaystyle\int_{x=0}^{x=\pi}\displaystyle\int_{y=1}^{y=2}x\sin(xy)dy \ dx &\quad &\text{Integrate first with respect to }y. \\ &=\displaystyle\int_{y=1}^{y=2}\displaystyle\int_{x=0}^{x=\pi}x\sin(xy)dx \ dy &\quad &\text{Integrate first with respect to }x. \end{align}[/latex]
b. If we want to integrate with respect to [latex]y[/latex] first and then integrate with respect to [latex]x[/latex], we see that we can use the substitution [latex]u = xy[/latex], which gives [latex]du = x\ dy[/latex]. Hence the inner integral is simply [latex]\displaystyle\int\sin u\ du[/latex] and we can change the limits to be functions of [latex]x,[/latex]
However, integrating with respect to [latex]x[/latex] first and then integrating with respect to [latex]y[/latex] requires integration by parts for the inner integral, with [latex]u = x[/latex] and [latex]dv = \sin(xy)dx[/latex].Then [latex]du = dx[/latex] and [latex]v =-\frac{\cos(xy)}{y}[/latex], so
