The single-variable definite integral measured the area under the graph of a function. Multivariable scalar fields also have graphs. For a scalar field of two variables, the graph is in \(\RR^3\) and there is volume under the graph instead of area. For scalar fields with three or more variables, the graphs are in higher dimensional spaces and enclose a hyper-volume. If the definite integral of a single variable function asked for the area under its graph, the definite integral for a multivariable integral should ask for the value of the volume or hyper-volume under its graph. This is, indeed, the case.
Let \(U\) be an subset of \(\RR^n\) and let \(f: U \rightarrow \RR\) be a scalar field. The graph of \(f\) is a hypersurface in \(\RR^{n+1}\) lying over \(U\) (thought of as a subset of the copy of \(\RR^n\) which is the \(x_{n+1} = 0\) hyperplane.) Then the integral
\begin{align*}
\int_U f dV \amp \amp \amp \int_U f(x,y) dA
\end{align*}
is the volume or hypervolume under the graph of \(f\) in \(\RR^{n+1}\text{.}\) If such a volume or hypervolume cannot be defined, then the integral does not exist. Like the single variable integral, (hyper)volume above the \(x_{n+1} =
0\) hyperplane is consider positive and (hyper)volume below is considered negative. The notation \(dV\) is used for scalar fields of three or more variables, indicating the (hyper)volume in the domain. For scalar fields of only two variables, the notation \(dA\) is used to represent the area in the domain.
Before moving on, I have a couple important notes to make about this definition. First, note that the scalar field needs to be defined on the set \(U\text{.}\) As with single variable integrals, I should always be aware of the domain of the scalar field. If the scalar field is undefined on some points in \(U\text{,}\) the integral will be undefined. If the scalar field is undefined on the boundary of \(U\text{,}\) the volume may still be defined. This will be an extension of improper integrals, which I will discuss in Subsection 7.1.1.
Also note that I said the integral is defined only if this volume is defined. Not all scalar fields will be integrable (indeed, not all single variable functions are integrable). Continuity will be enough to conclude this volume exists, but I will also investigate less restrictive conditions when I define the Riemann integral below. In addition, not all subsets \(U\) allow for this volume to be defined. The definitions in Section 6.4 will describe which sets \(U\) lead to well-defined volumes. For the integrals to be defined, both the scalar field \(f\) and the subset \(U\) have to be reasonable.
Finally, note that this is a very conceptual definition, but it is hard to work with. I will try to work towards a better definition that allows for calculation. I can’t start with arbitrary domains \(U\text{;}\) I need to start with simpler domains: intervals.
Now I’m going to try to construct the definite integral extending the Riemann integral from single variable calculus. Recall that definition, noting that it was explicity defined for a function whose domain was an interval.
Let \(f : [a,b] \rightarrow \RR\) be a function. The Riemann integral of \(f\) on the interval \([a,b]\) is the following limit (if it exists), where in the sum the interval is divided into \(n\) equal pieces and \(x_k^*\) is any point in the \(k\)th subinterval.
I want to extend the pieces of this definition, now that I had defined the multivariable interval. For the single-variable integral, I partitioned the interval into \(n\) pieces. I can also partition these multi-variable intervals. In general, there are several ways to do so; one of the easiest is simply to divide each of the single variable intervals into \(k\) pieces. In \(\RR^2\text{,}\) this splits the rectangle into a checkerboard pattern, with \(k^2\) small rectangles. In \(\RR^3\text{,}\) it splits into \(k^3\) small rectangular prisms. In \(\RR^n\) there are \(k^n\) subdivisions.
In a partition of \(I\) into \(k^n\) pieces, I can use the same limit process as the single variable integral. Let \(x^*\) be any point in each piece of the partition and estimate the (hyper-)volume over that piece of the interval by a rectangular (hyper-)volume. The height is \(f(x^*)\text{.}\) The (hyper-) volume of one pieces of the approximation is this height multiplied by the size of the small partition pieces, in the appropriate dimension. I will write \(\Delta V\) for this size, thinking of \(V\) for volume or hypervolume. (As before, in the special case of two variables, I would write \(\Delta A\) instead). I’ll define this for a three-variable scalar field for clarity of notation, but the definition the same for any number of variables.
Let \(I \subset \RR^3\) be an interval and \(f: I
\rightarrow \RR\) a scalar field. The Riemann integral of a scalar field is the following limit (if it exists), where the sum is taken over all \(n^3\) subdivision of the interval and \(x_k^*\) is any point in the \(k\)th subinterval (by some way of counting all the intervals), and \(\Delta V\) is the (hyper)volume of the subinterval.
In the multivariable integral, the \(\Delta V\) becomes an infinitesimal piece of area, volume or hypervolume, which explains the notation \(dV\) that I used in Definition 6.3.1.
In Definition 6.3.3, I defined an integrable scalar field, on an interval, to be a scalar field such that the limit defining the Riemann integral exists. Most usual scalar fields are integrable, particularly scalar fields defined on closed intervals (so that they don’t diverge near the edges of the interval). There is a whole theory of integrable higher dimensional functions. However, for the purposes of this course, I’m going to work with a slightly restricted class of scalar fields. This won’t affect any applications, since all the reasonable scalar fields I want to use will still be in this category. The non-integrable functions are, for practical purposes, weird inventions by mathematics to try to break our own definitions (which is, after all, a think we just like to do).
Let \(I\) be a closed interval in \(\RR^n\) and let \(f: I \rightarrow \RR\) be a scalar field. If I can break the interval \(I\) up into finitely many closed subintervals and if \(f\) is continuous on each of the closed subintervals, I say the scalar field is piecewise continuous.
As I said above, this is a convenience, since piecewise continuous is good enough for our purposes. There are, in fact, other scalar fields which are also integrable. Notice, also, that I defined this exlictly on closed intervals. Piecewise-continuous scalar fields on open intervals can have divergent behaviour (like vertical asymptote) at the edges of the intervals. These produce a multivariable version of improper integrals. I will consider these later, but insisting on scalar fields defined on closed intervals ensures that I don’t have problems like asymptotes at the edges.
As an aside, there are also different notions of integrability and different definition of the integral. What I have discussed here is the Riemann integral, and this notion of integrability is Riemann-integrability. In higher mathematics, the most commonly used definition of the integral is a different definition, called the Lebesgue integral. It is also an approximation process, but it involves approximating the function with a special functions that only take finitely many values. The Lebesgue integral agrees with the Riemann integral on all Riemann-integrable functions. However, there are functions where the Lebesgue integral is defined but the limit in the Riemann integral fails to converge. It is interesting to observe that something as fundamental as the integral has multiple definitions and the choice of definitions has a real, measurable effect on our mathematics. A different kind of calculus course, one more interested in the mathematical foundations, would delve into the details of these different notions of the integral and integrebility.
Let \(I\) be an interval in \(\RR^n\text{,}\)\(f: I
\rightarrow \RR\) and \(g: I \rightarrow \RR\) integrable scalar fields on I, and \(c \in \RR\text{.}\) First, the integral is linear.
\begin{equation*}
\int_I f dV \leq \int_I g dV
\end{equation*}
Let \(J\) be another interval where \(f\) is integrable and assume \(J\) is of matching size and adjacent to \(I\) (such that they have no open intersection and their union \(I \cup J\) is also also an interval).
