Double Integrals

Definition and Properties of Double Integrals

Definition of Double Integral

The definite integral can be extended to functions of more than one variable. Consider, for example, a function of two variables \(z = f\left( {x,y} \right).\) The double integral of function \(f\left( {x,y} \right)\) is denoted by
\[\iint\limits_R {f\left( {x,y} \right)dA},\] where \(R\) is the region of integration in the \(xy\)-plane.

A function of two variables z=f(x,y) over region R

Figure 1.

If the definite integral \(\int\limits_a^b {f\left( x \right)dx} \) of a function of one variable \({f\left( x \right)} \ge 0\) is the area under the curve \({f\left( x \right)}\) from \(x = a\) to \(x = b,\) then the double integral is equal to the volume under the surface \(z = f\left( {x,y} \right)\) and above the \(xy\)-plane in the region of integration \(R\) (Figure \(1\)).

As in the case of integral of a function of one variable, a double integral is defined as a limit of a Riemann sum.

Partition of a rectangular region of integration into small intervals

Figure 2.

If the region \(R\) is a rectangle \(\left[ {a,b} \right] \times \left[ {c,d} \right]\) (Figure \(2\)), we can subdivide \(\left[ {a,b} \right]\) into small intervals with a set of numbers
\(\left\{ {{x_0},{x_1}, \ldots ,{x_m}} \right\}\) so that
{a }={ {x_0} \lt {x_1} \lt {x_2} \lt \ldots} \lt{ {x_i} \lt \ldots }
\lt { {x_{m – 1}} \lt {x_m} }={ b.}
\] Similarly, a set of numbers \(\left\{ {{y_0},{y_1}, \ldots ,{y_n}} \right\}\) is said to be a partition of \(\left[ {c,d} \right]\) along the \(y\)-axis, if
{c }={ {y_0} \lt {y_1} \lt {y_2} \lt \ldots} \lt{ {y_j} \lt \ldots }
\lt { {y_{n – 1}} \lt {y_n} }={ d.}

The Riemann sum of a function f (x,y) over this partition of \(\left[ {a,b} \right] \times \left[ {c,d} \right]\) is
\[\sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n {f\left( {{u_i},{v_j}} \right)\Delta {x_i}\Delta {y_j}} } ,\] where \({\left( {{u_i},{v_j}} \right)}\) is some point in the rectangle \(\left( {{x_{i – 1}},{x_i}} \right) \) \(\times \left( {{y_{j – 1}},{y_j}} \right)\) and \(\Delta {x_i} = {x_i} – {x_{i – 1}},\) \(\Delta {y_j} = {y_j} – {y_{j – 1}}.\)

We then define the double integral of a function \({f\left( {x,y} \right)}\) in the rectangular region \(\left[ {a,b} \right] \times \left[ {c,d} \right]\) to be the limit of the Riemann sum as maximum values of \(\Delta {x_i}\) and \(\Delta {y_j}\) approach zero:
{\iint\limits_{\left[ {a,b} \right] \times \left[ {c,d} \right]} {f\left( {x,y} \right)dA} }
= {\lim\limits_{\substack{
\text{max}\,\Delta {x_i} \to 0\\
\text{max}\,\Delta {y_j} \to 0}} \sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n {f\left( {{u_i},{v_j}} \right)}}}\kern0pt{{{\Delta {x_i}\Delta {y_j}} } .}

To define the double integral over a bounded region \(R\) other than a rectangle, we choose a rectangle \(\left[ {a,b} \right] \times \left[ {c,d} \right]\) that contains \(R\) (Figure \(3\text{),}\) and we define the function \({g\left( {x,y} \right)}\) so that
{g\left( {x,y} \right) = f\left( {x,y} \right), \;}\kern-0.3pt {\text{if}\;\;f\left( {x,y} \right) \in R }\\
{g\left( {x,y} \right) = 0, \;}\kern-0.3pt {\text{if}\;\;f\left( {x,y} \right) \notin R}
\] Then, the double integral of the function \({f\left( {x,y} \right)}\) over a general region \(R\) is defined to be
{\iint\limits_R {f\left( {x,y} \right)dA} }
= {\iint\limits_{\left[ {a,b} \right] \times \left[ {c,d} \right]} {g\left( {x,y} \right)dA}.}

Elementary general region of integration in double integral

Figure 3.

Region S in region R in double integral

Figure 4.

Union of regions R and S in double integral

Figure 5.

Properties of Double Integrals

The double integral satisfies the following properties:

  1. \({\iint\limits_R {\left[ {f\left( {x,y} \right) + g\left( {x,y} \right)} \right]dA} }\) \(= {\iint\limits_R {f\left( {x,y} \right)dA} }\) \(+{ \iint\limits_R {g\left( {x,y} \right)dA} ;}
  2. \({\iint\limits_R {\left[ {f\left( {x,y} \right) – g\left( {x,y} \right)} \right]dA} }\) \(= {\iint\limits_R {f\left( {x,y} \right)dA} }\) \(-{ \iint\limits_R {g\left( {x,y} \right)dA} ;}
  3. \(\iint\limits_R {kf\left( {x,y} \right)dA} \) \( = k\iint\limits_R {f\left( {x,y} \right)dA},\) where \(k\) is a constant;
  4. If \({f\left( {x,y} \right)} \le {g\left( {x,y} \right)}\) on \(R,\) then \(\iint\limits_R {f\left( {x,y} \right)dA} \) \(\le \iint\limits_R {g\left( {x,y} \right)dA} ;\)
  5. If \({f\left( {x,y} \right)} \ge 0\) on \(R\) and \(S \subset R\) (Figure \(4\)), then \(\iint\limits_S {f\left( {x,y} \right)dA} \) \(\le \iint\limits_R {f\left( {x,y} \right)dA} ;\)
  6. If \({f\left( {x,y} \right)} \ge 0\) on \(R\) and \(R\) and \(S\) are non-overlapping regions (Figure \(5\)), then \(\,{\iint\limits_{R \cup S} {f\left( {x,y} \right)dA} }\) \(= {\iint\limits_R {f\left( {x,y} \right)dA} }\) \(+{ \iint\limits_S {f\left( {x,y} \right)dA} .} \)
    Here \({R \cup S}\) is the union of these two regions.

Solved Problems

Click on problem description to see solution.

Example 1.

Let \(R\) and \(S\) be non-overlapping regions (Figure \(5\)). The values of double integrals are known:
{\iint\limits_R {f\left( {x,y} \right)dA} = 2,\;\;}\kern-0.3pt
{\iint\limits_R {g\left( {x,y} \right)dA} = 3,\;\;}\kern-0.3pt
{\iint\limits_S {f\left( {x,y} \right)dA} = 6,\;\;}\kern-0.3pt
{\iint\limits_S {g\left( {x,y} \right)dA} = 7.}
\] Evaluate the integral \(\iint\limits_{R \cup S} {\left[ {10f\left( {x,y} \right) }\right.}\) \(+\,{\left.{ 20g\left( {x,y} \right)} \right]dA} .\)


Using properties of the double integrals, we have
{\iint\limits_{R \cup S} {\left[ {10f\left( {x,y} \right) }\right.}+{\left.{ 20g\left( {x,y} \right)} \right]dA} }
= {\iint\limits_{R \cup S} {10f\left( {x,y} \right)dA} }+{ \iint\limits_{R \cup S} {20g\left( {x,y} \right)dA} }
= {{10\iint\limits_{R \cup S} {f\left( {x,y} \right)dA} }+{ 20\iint\limits_{R \cup S} {g\left( {x,y} \right)dA} }}
= {{10\left[ {\iint\limits_R {f\left( {x,y} \right)dA} }\right.}+{\left.{ \iint\limits_S {f\left( {x,y} \right)dA} } \right] }}
+ {{20\left[ {\iint\limits_R {g\left( {x,y} \right)dA} }\right.}+{\left.{ \iint\limits_S {g\left( {x,y} \right)dA} } \right] }}
= {10\left( {2 + 6} \right) }+{ 20\left( {3 + 7} \right) }={ 280.}