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Calculus 3 (Math 221) Notes and Study Guide

Brenton LeMesurier

Section 5.2 Double Integrals over General Regions

Corrections made on 2025-04-26

References.

Topics.

Subsection 5.2.1 The Double Integral Over a Bounded Domain

Consider the problem of calculating the volume of the solid that sits above the bounded region \(D\) in the plane, and under the surface \(z=f(x,y)\text{.}\) Since the region is bounded, one can surround it with a rectangle \(R\text{,}\) and then extend the solid with a zero volume sheet covering the part of \(R\) outside \(D\text{.}\)
The resulting solid is the one over \(R\) and under the graph of the extended function
\begin{equation*} F(x,y) = \left\{ \begin{array}{ll} f(x,y) \amp \text{ for } (x,y) \in D, \\ 0 \amp \text{ for } (x,y) \text{ in } R \text{ but not in } D. \end{array} \right. \end{equation*}
Thus it seems reasonable to define the volume of this solid as \(\iint_R F(x,y)\ dA\text{,}\) and to define the double integral of \(f\) over \(D\) as the integral of this extended function over the extended domain \(R\text{:}\)

Definition 5.2.1. Integral over a bounded domain \(D\).

For any rectangle \(R=[a,b]\times[c,d]\) containing bounded region \(D\)
\begin{equation} \iint\limits_D f(x,y)\ dA := \iint\limits_R F(x,y)\ dA\tag{5.2.1} \end{equation}
(Note the notation "\(A := B\)" for "\(A\) is defined to be \(B\)".)

Does This Integral Exist?

However, at the boundary of the domain, \(F\) is likely to be discontinuous, jumping to value zero outside \(D\text{,}\) so does this integral even exist?
Fubini’s Theorem gives an answer: so long as the boundary of \(D\) is made up of a finite set of smooth curves, \(F\) is continuous everywhere except on those curves, and that is enough for its double integral to exist (and to be computable using iterated integrals.)

Subsection 5.2.2 Iterated Integrals over Non-rectangular Regions

Type I Regions.

Double integrals are easiest when the domain can be described by inequalities such as a domain of Type I (which I also like to call "Type \(dy\ dx\)", because as we will see, that indicates the order in which the integrations are done):
\begin{equation*} D = \{(x,y)| a \leq x \leq b, B(x) \leq y \leq T(x) \} \end{equation*}
This region is bounded above and below by curves \(y=T(x)\text{,}\) \(y=B(x)\text{,}\) and at left and right by vertical lines \(x=a\) and \(x=b\text{.}\)
Thus the domain divides into a collection of vertical line segments, and the integral can be done first along each such line (integral in \(y\) with fixed \(x\)) and then integrating the resulting value over \(x\text{:}\)
\begin{align} \iint\limits_D f(x,y)\ dA \amp= \int\limits_{a}^b \int\limits_{B(x)}^{T(x)} f(x,y)\ dy\ dx\notag\\ \amp= \int\limits_{x=a}^b \left[ \int\limits_{y=B(x)}^{T(x)} f(x,y)\ dy \right] dx\tag{5.2.2} \end{align}

Type II Regions.

Reversing the roles of \(x\) and \(y\) gives a domain of Type II (which I also call "Type \(dx\ dy\)"):
\begin{equation*} D = \{(x,y)| c \leq y \leq d, L(y) \leq x \leq R(y) \} \end{equation*}
This is the collection of horizontal lines bounded below by \(y=c\text{,}\) above by \(y=d\text{,}\) to the left by curve \(x=L(y)\text{,}\)and to the right by \(x=R(y)\text{,}\) and the integral can be evaluated as
\begin{equation} \iint\limits_D f(x,y) dA = \int\limits_{y=c}^d \left[ \int\limits_{x=L(y)}^{R(y)} f(x,y)\ dx \right] dy\tag{5.2.3} \end{equation}

Subsection 5.2.3 A Strategy for Evaluating Double Integrals

In setting up double integrals as iterated integrals, the first question to ask is:
Can the minimum and maximum allowed values of one variable be specified as functions of the other?
If so, the integral is done over the variable whose values are so described, and then over the other variable.
The integral done last must have numerical lower and upper limits, not depending on another variable.

Remark 5.2.4.

Note well: If the upper or lower limit on one of the iterated integrals depends on another dummy variable of integration, this integral must be inside the integral over that other variable, because that other dummy variable only has a specific variable inside the integral over that variable.

Subsection 5.2.4 Properties of Double Integrals

The three properties stated in Subsection 5.1.2 for double integrals over a rectangle apply to all double integrals; they are worth updating here for the more general case, and adding some more.
  1. \begin{equation} \iint\limits_D f(x,y) + g(x,y) \ dA = \iint\limits_D f(x,y) \ dA + \iint\limits_D g(x,y) \ dA.\tag{5.2.4} \end{equation}
  2. \begin{equation} \iint\limits_D k f(x,y) ) \ dA = k \iint\limits_D f(x,y) \ dA \text{ for any } k.\tag{5.2.5} \end{equation}
  3. \begin{equation} \text{ If } f(x,y) \leq g(x,y)\text{ then } \iint\limits_D f(x,y) ) \ dA \leq \iint\limits_D g(x,y) \ dA\tag{5.2.6} \end{equation}
    when the first inequality holds for all points \((x,y)\) in \(D\text{.}\)
  4. Further, a domain can be cut into pieces and the integrals over the pieces added:
There is now a natural definition of the area of a region \(D\) as a double integral, by identifying it as the volume of the cylinder of height 1 over that region:

Definition 5.2.6.

The area of region \(D\) is
\begin{equation} A(D) = \iint\limits_D 1\ dA\tag{5.2.8} \end{equation}

Exercise 5.2.7.

Verify that this gives the area of the rectangular region \(R = [a,b] \times [c,d]\) to be \(A(R)=(b-a)(d-c)\text{.}\)
The easy way to do this is via a simple iterated integral, but as a further exercise one can also calculate this using the original definition in terms of limits of sums, Subsection 5.1.1.

Exercise 5.2.8.

  1. Sketch the triangular region between the lines \(y=0\text{,}\) \(x=3\) and \(y = 2x\)
  2. Show that it is both of Type I and Type II, and set up iterated integral expressions for its area in the corresponding "\(dx\ dy\)" and "\(dy\ dx\)" forms.
  3. Use one or both of these integrals to calculate its area.
  4. Check your answer with geometry.
Combined with the inequality of Double Integral Property 3 in (5.2.6),

Subsection 5.2.5 Domains of More Complicated Shapes: Divide and Conquer

Not all domains fit the types seen so far.
For other domain shapes, often the best hope is to seek a way to divide the domain into several pieces, each of one of the types above. For a region whose boundary consists of one or more smooth curves, vertical cuts through each point on the boundary with a vertical tangent will (usually) divide it into a collection of Type I regions. (Likewise cutting on each horizontal tangent will divide it into Type II regions, and sometimes a mixture of both can give simpler pieces.)
Then the total integral is the sum of the integral over these pieces, using Double Integral Property 4 as in Equation (5.2.7).
Sometimes, the formulas for the curves at the top and bottom (and/or at left and right) change at certain points; then a few extra vertical (and/or horizontal) cuts at these transition points give smooth curves and a single formula for the limits on each piece.
This strategy is not quite universal —it assumes for example that there are only a finite number of boundary points with vertical (resp. horizontal) tangents— but it is widely useful.

Exercise 5.2.10.

Draw some regions with "meandering boundaries" (not of either Type I or Type II) and see what these vertical and/or horizontal tangent line cuts do.

Subsection 5.2.6 Changing the Order of Integration

Unlike with double integrals over rectangles, one cannot simply swap the order of the integrals in an iterated integral when the limits of the inner integral depend on the variable in the outer one. For example, the upper half of the unit disk is of both Type I and Type II and an integral over it is given by
\begin{equation} \int_{x=-1}^1 \int_{y=0}^{\sqrt{1-x^2}}\ dy\ dx\tag{5.2.10} \end{equation}
but swapping those limits gives
\begin{equation*} \int_{y=0}^{\sqrt{1-x^2}} \int_{x=-1}^1\ dx\ dy \end{equation*}
which is nonsensical, because the limits on the outer integral cannot depend on the "inner", dummy variable \(x\text{.}\)
Instead, swapping requires several steps:
  1. Check that the domain is of both Type I and Type II; if not, divide it into pieces tht are.
  2. Work out the full range of possible values that can be attained by the inner variable over all possible values of the outer variable, by looking at the extreme possible values, which correspond to values on the boundary of the domain: these become the limits of the new outer integral.
  3. For each value of this new outer integral variable, solve for the allowable values of the new inner one: these will now (in general) depend on the new outer variable, and become the limits of the new inner integral.
For this, it helps to express the domain in terms of inequalities, and then solve the corresponding equations to descibe the boundary of the domain: those equations describe the relation between \(x\) and \(y\) that applies at the two limits of the inner integral.
It can also be very helpful to sketch the domain!

Example 5.2.11.

For an integral over the upper half of ther unit disk, the Type I form is as above in (5.2.10), and the domain is
\begin{equation*} D = \{(x,y) | -1 \le x \le 1, 0 \leq y \le \sqrt{1-x^2}\} \end{equation*}
(Note that the variables are always listed "from the outside inward".)
Considering all allowable value for \(x\) the highest allowable value of \(y\) is \(1\text{,}\) occuring for \(x = \pm 1\) and the lowest value is simply \(0\text{.}\)
Thus \(0 \leq y \leq 1\) is the overall range, giving the outer integral \(\int_{y=0}^1 \dots dy\)
Next, the boundary consists of all points at the limits of the inner integral:
\begin{equation*} y = 0 \text{ and } y = \sqrt{1-x^2} \end{equation*}
Each of these can be solved for \(x\text{:}\)
  1. \(y=0\) adds no constraints on \(x\) other than the overall restriction \(-1 \le x \le 1\text{.}\)
  2. \(y=\sqrt{1-x^2}\) gives the extremities \(x^2 = 1-y^2\) and thus the solutions \(x = -\sqrt{1-y^2}\) and \(x = \sqrt{1-y^2}\text{,}\) limiting points in the domain to \(-\sqrt{1-y^2} \le x \le \sqrt{1-y^2}\text{.}\)
  3. The combination of these restriction is just the latter, so for a given value of \(y\text{,}\) the values of \(x\) in the domain are \(-\sqrt{1-y^2} \le x \le \sqrt{1-y^2}\)
  4. That is, the domain is
    \begin{equation*} D = \{(x,y)| -1 \le y \le 1, -\sqrt{1-y^2} \leq x \le \sqrt{1-y^2}\} \end{equation*}
    which means that the Type II iterated integral form (slicing the half-disk horizontally) is
    \begin{equation} \int_{y=0}^1 \int_{x=-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\ dx\ dy\tag{5.2.11} \end{equation}

Exercise 5.2.12.

Make sketches to illustrate this: first the vertical slicing of Type I and then the horizontal slicing of Type II.

Example 5.2.13.

The iterated integral
\begin{equation} \int_{y=0}^1 \int_{x=y}^1 \sin(x^2)\ dx\ dy\tag{5.2.12} \end{equation}
can not be evaluated in this order in terms of elementary functions, because there is no elementary anti-derivative of \(\sin(x^2)\text{.}\) However, it can be evaluated by first changing the order of integration.
Firstly, \(y \leq x \leq 1\) and \(0 \leq y \leq 1\) gives the range of all possible values of \(x\) as \(0 \leq x \leq 1\text{.}\)
Then for a given value of \(x\text{,}\) the extremes of \(y\) come from \(y \leq x\) and \(y=0\text{:}\) that is, \(0 \leq y \leq x\text{,}\) giving the new iterated integral form
\begin{equation} \int_{x=0}^1 \int_{y=0}^x \sin(x^2)\ dy\ dx, = \int_{x=0}^1 \left[ ]\int_{y=0}^x\ dy \right] \sin(x^2)\ dx.\tag{5.2.13} \end{equation}
The inner integral is now easy:
\begin{equation*} \int_{y=0}^x \ dy = \Big[ y \Big]_{y=0}^x = [x - 0] = x, \end{equation*}
so the iterated integral (5.2.13) gives
\begin{equation*} \int_{x=0}^1 x \sin(x^2)\ dx \end{equation*}
Finally, the substitution \(u=x^2\text{,}\) \(du = 2 x dx\) gives
\begin{equation*} \int_{u=0}^1 \sin(u) \frac{du}{2} = \left[ \frac{-\cos(u)}{2} \right]_{u=0}^1 = \frac{-\cos(1) -(-\cos(0))}{2} = \frac{1 - \cos(1)}{2}. \end{equation*}

Subsection 5.2.7 The Average Value of a Function over a Region

The average value of a function \(f(x)\) over interval \([a,b]\) was defined as the definite integral over that interval divided by the "size" of the interval, measured by its length \(b-a\text{.}\) This is the natural choice of size, because it ensures that the average of a constant function equals its value.
Similarly, it makes sense to define the average value of a function over region \(R\) as the integral divided by the area \(A(R)\) of the region:
\begin{equation} f_{ave} = \overline{f} = \frac{\iint\limits_R f(x,y)\ dA}{A(R)} = \frac{\iint\limits_R f(x,y)\ dA}{\iint\limits_R \ dA}\tag{5.2.14} \end{equation}
For the case of rectangle \(R=[a,b] \times [c,d]\text{,}\) \(A(R)=(b-a)(d-c)\text{,}\) so
\begin{equation*} f_{ave} = \overline{f} = \frac{\iint\limits_R f(x,y)\ dA}{(b-a)(d-c)}. \end{equation*}

Subsection 5.2.8 Optional Topic: Improper Double Integrals

We can make sense of an integral over an infinite "corner" region like
\begin{equation*} D = \{(x,y)| a \leq x \le \infty , c \leq y \le \infty \} \end{equation*}
as a double limit in both the upper limits:
\begin{align} \iint\limits_D f(x,y)\ dA \amp= \lim\limits_{(b,d) \to (\infty, \infty)} \int\limits_{x=a}^b \int\limits_{y=c}^d f(x,y)\ dy\ dx\tag{5.2.15}\\ \amp= \lim\limits_{(b,d) \to (\infty, \infty)} \int\limits_{y=c}^d \int\limits_{x=a}^b f(x,y)\ dx\ dy\tag{5.2.16} \end{align}

Example 5.2.14.

For domain \(D = [0,\infty) \times [0,\infty)\)
\begin{gather*} \iint_D x y e^{-(x^2 + y^2)} dA = \lim\limits_{(b,d) \to (\infty, \infty)} \int_{x=0}^b \int_{y=0}^d x y e^{-(x^2 + y^2)} \ dy\ dx\\ = \left( \lim_{b \to \infty} \int_0^b x e^{-x^2}\ dx \right) \left(\lim_{d \to \infty} \int_0^d y e^{-y^2}\ dy \right) = \left( \lim_{b \to \infty} \int_0^b x e^{-x^2}\ dx \right)^2 \end{gather*}
The single integral can be evaluated using the substitution \(u = x^2\text{,}\) \(du = 2x dx\) as
\begin{equation*} \frac{1}{2} \lim_{B \to \infty} \int_0^B e^{-u}\ du = \frac{1}{2} \lim_{B \to \infty}[-e^{-u}]_{u=0}^B = \frac{1}{2} \lim_{B \to\infty} (-e^{-B} + e^{-0}) = 1/2, \end{equation*}
so the double integral is \(\displaystyle \iint_D x y e^{-(x^2 + y^2)} dA = 1/4.\)

Study Guide.

Study Section 5.2 of OpenStax Calculus Volume 3
 9 
openstax.org/books/calculus-volume-3/pages/5-2-double-integrals-over-general-regions
up to the subsection on Calculating Volumes, Areas, and Average Values; that is, omitting the final topic of Improper Double Integrals (However, study it if interested: such integrals come up in physics and mathematical probability.) In particular
  • Theorems 3 to 7.
  • The Definitions of Type I and Type II regions, and how to visualize them.
  • The strategy for changing the order of integration, by converting between a "Type I description" and a "Type II description".
  • Examples 11–18, and the Checkpoints following each.
  • Exercises 61–63 (They go together.)
  • Exercises 66 and 67. (They go together.)
  • Exercise 72 or 73.
  • One or more of Exercises 74 to 77.
  • Exercise 78 and/or 79.
  • One or more of Exercises 80 to 85.
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