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Section 3.1 Multiple Integrals

In one-variable calculus, the integral of a function over an interval measures signed area. In several variables, the same idea measures signed area, volume, or higher-dimensional content. We will not develop the general theory in detail. Instead, we will assume that the functions and domains we consider are integrable.

Definition 3.1.1. Integral over a Domain.

Let \(D \subseteq \R^n\) be a domain, and let \(f \colon D \to \R\text{.}\) Partition \(D\) into small pieces \(D_1,\ldots,D_N\text{,}\) choose sample points \(\vx_j \in D_j\text{,}\) and form the Riemann sum
\begin{equation*} \sum_{j=1}^N f(\vx_j)\operatorname{vol}(D_j). \end{equation*}
If these Riemann sums approach a common limit as the partition is refined and the pieces become arbitrarily small, then \(f\) is called integrable on \(D\text{,}\) and that limit is the integral of \(f\) over \(D\text{,}\) denoted by
\begin{equation*} \int_D f(\vx)\,dV. \end{equation*}
This is also called a multiple integral.
The symbol \(dV\) indicates integration with respect to volume in \(\R^n\text{.}\) When \(f(\vx)=1\text{,}\) the integral \(\int_D 1\,dV\) is the \(n\)-dimensional volume of \(D\text{.}\)

Definition 3.1.2. Double and Triple Integrals.

When \(D \subseteq \R^2\text{,}\) we usually write
\begin{equation*} \iint_D f(x,y)\,dA \end{equation*}
and call it a double integral. When \(E \subseteq \R^3\text{,}\) we usually write
\begin{equation*} \iiint_E f(x,y,z)\,dV \end{equation*}
and call it a triple integral.
The notation reminds us of the dimension of the domain, but conceptually these are all the same operation: integrating a scalar field over a region.
The coordinate axes and a shaded rectangular region in the plane with corners labeled by.
Figure 3.1.3. A double integral over a rectangle.
The main use of Fubini’s theorem is computational. It reduces a multiple integral to repeated one-variable integrals. In practice, this is how most double and triple integrals are evaluated.

Example 3.1.5.

Compute
\begin{equation*} \iint_R (x+2y)\,dA, \qquad R=[0,1]\times[1,3]. \end{equation*}
Solution.
By Fubini’s theorem,
\begin{equation*} \iint_R (x+2y)\,dA = \int_0^1 \int_1^3 (x+2y)\,dy\,dx. \end{equation*}
Integrating first with respect to \(y\text{,}\)
\begin{equation*} \int_1^3 (x+2y)\,dy = xy+y^2\Big|_{y=1}^{y=3} = 2x+8. \end{equation*}
Therefore
\begin{equation*} \iint_R (x+2y)\,dA = \int_0^1 (2x+8)\,dx = x^2+8x\Big|_{0}^{1} = 9. \end{equation*}
For nonrectangular regions, one usually describes the domain by inequalities. For example, if
\begin{equation*} D=\{(x,y): a \le x \le b,\ \phi_1(x) \le y \le \phi_2(x)\}, \end{equation*}
then
\begin{equation*} \iint_D f(x,y)\,dA = \int_a^b \int_{\phi_1(x)}^{\phi_2(x)} f(x,y)\,dy\,dx. \end{equation*}
This corresponds to slicing the region by vertical line segments. One can also slice horizontally when that description is more convenient.
The shaded region in the plane between the curves y equals x squared and y equals x for x.
Figure 3.1.6. A region described by vertical slices.

Example 3.1.7.

Let
\begin{equation*} D=\{(x,y): 0 \le x \le 1,\ x^2 \le y \le x\}. \end{equation*}
Compute
\begin{equation*} \iint_D x\,dA. \end{equation*}
Solution.
The region is described by vertical slices, so
\begin{equation*} \iint_D x\,dA = \int_0^1 \int_{x^2}^{x} x\,dy\,dx. \end{equation*}
Since \(x\) is constant with respect to \(y\text{,}\)
\begin{equation*} \int_{x^2}^{x} x\,dy = x(y)\Big|_{y=x^2}^{y=x} = x(x-x^2) = x^2-x^3. \end{equation*}
Hence
\begin{equation*} \iint_D x\,dA = \int_0^1 (x^2-x^3)\,dx = \frac13-\frac14 = \frac{1}{12}. \end{equation*}

Example 3.1.8.

Compute
\begin{equation*} \iiint_B (x+y+z)\,dV, \qquad B=[0,1]\times[0,2]\times[0,3]. \end{equation*}
Solution.
We may integrate in the order \(dz\,dy\,dx\text{:}\)
\begin{equation*} \iiint_B (x+y+z)\,dV = \int_0^1 \int_0^2 \int_0^3 (x+y+z)\,dz\,dy\,dx. \end{equation*}
The inner integral is
\begin{equation*} \int_0^3 (x+y+z)\,dz = 3x+3y+\frac92. \end{equation*}
Therefore
\begin{equation*} \iiint_B (x+y+z)\,dV = \int_0^1 \int_0^2 \left(3x+3y+\frac92\right)\,dy\,dx. \end{equation*}
Integrating with respect to \(y\) gives
\begin{equation*} \int_0^2 \left(3x+3y+\frac92\right)\,dy = 6x+15. \end{equation*}
Finally,
\begin{equation*} \iiint_B (x+y+z)\,dV = \int_0^1 (6x+15)\,dx = 18. \end{equation*}
The essential idea is always the same: describe the domain in a useful way, choose an order of integration, and then compute the resulting iterated integrals one variable at a time.
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