Double Integrals Over Rectangles

The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, $f(x, y)$ or $f(x, y, z)$. Integrals of a function of two variables over a region in $R^2$ are called double integrals. Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the $x$-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three dimensional Cartesian plane where $z = f(x, y))$ and the plane which contains its domain. The same volume can be obtained via the triple integral—the integral of a function in three variables—of the constant function $f(x, y, z) = 1$ over the above-mentioned region between the surface and the plane. If there are more variables, a multiple integral will yield hypervolumes of multi-dimensional functions.

Volume to be Integrated

Double integral as volume under a surface $z = x^2 − y^2$. The rectangular region at the bottom of the body is the domain of integration, while the surface is the graph of the two-variable function to be integrated.

Double Integrals Over Rectangles

Double integrals over rectangular regions are straightforward to compute in many cases. For a rectangular region $S$ defined by $x$ in $[a,b]$ and $y$ in $[c,d]$, the double integral of a function $f(x,y)$ in this region is given as:

$\begin{aligned}\int\!\!\!\int_S f(x,y) dxdy &= \int_a^b\left(\int_c^d f(x,y) dy\right) dx \\ &= \int_c^d\left(\int_a^b f(x,y) dx\right) dy\end{aligned}$.

Here, we exchanged the order of the integration, assuming that $f(x,y)$ satisfies the conditions to apply Fubini's theorem.

Example

Let us assume that we wish to integrate a multivariable function $f$ over a region $A$:

$A = \left \{ (x,y) \in \mathbf{R}^2 : 11 \le x \le 14 \ ; \ 7 \le y \le 10 \right \}$

$f(x,y) = x^2 + 4y$

Formulating the double integral , we first evaluate the inner integral with respect to $x$:

$\begin{aligned} \int_{11}^{14} (x^2 + 4y) \ dx & = \left (\frac{1}{3}x^3 + 4yx \right)\Big |_{x=11}^{x=14} \\ & = \frac{1}{3}(14)^3 + 4y(14) - \frac{1}{3}(11)^3 - 4y(11) \\ &= 471 + 12y \end{aligned}$

We then integrate the result with respect to $y$:

$\begin{aligned} \int_7^{10} (471 + 12y) \ dy & = (471y + 6y^2)\big |_{y=7}^{y=10} \\ & = 471(10)+ 6(10)^2 - 471(7) - 6(7)^2 \\ &= 1719 \end{aligned}$

We could have computed the double integral starting from the integration over $y$. Confirm yourself that the result is the same.