Single-variable definite integrals are first used to compute the area bounded by a set of curves. The integral has a multitude of applications, but computing a planar area is one of the best visual ways to understand what it means.

It makes sense that the best way to introduce definite integrals of many-variable functions is through *their* most visual application: computing volumes of 3D regions bounded by surfaces.

We'll use the Riemann sum here to make the jump from single-variable integrals to **double integrals**, covered in much greater detail in the Multiple Integrals chapter.

A classic problem from single-variable calculus is to compute the area bounded above by the parabola $y = x^2,$ below by the $x$-axis, and between the lines $x = 0$ and $x =1.$

In short, we're going to compute the **definite integral**
$\int\limits_{x=0}^{x=1} x^2 dx .$
The rules of antidifferentiation and the fundamental theorem of calculus gives us the answer in a short calculation. We're not so much interested in the value here as we are in the process.

A definite integral is really the limit of a Riemann sum: $\int\limits_{x=a}^{x=b} f(x)\, dx =\lim\limits_{n \to \infty } \sum\limits_{j=1}^{n} f(x_{j}) \left( \frac{b-a}{n} \right).$ If we break the interval $[a,b]$ up into $n$ equal-sized pieces, $x_{j}$ is the right-hand endpoint of the $j^\text{th}$ subinterval.

What's a general formula for $x_{j}?$

We use thin rectangles of height $f \left(a+ j \left[ \frac{b-a}{n} \right] \right)$ and base length $\frac{b-a}{n}$ to estimate the area of a planar region.

As $n \to \infty$ and more and more rectangles are used, the area approximation gets better and better.

The limit (when it exists) is the definite integral $\int\limits_{x=a}^{x=b} f(x) dx =\lim\limits_{n \to \infty } \sum\limits_{j=1}^{n} \underbrace{f \left(a+ j \left[ \frac{b-a}{n} \right] \right)}_{\text{height}} \underbrace{\left( \frac{b-a}{n} \right)}_{\text{base}}.$ Compute $\int\limits_{x=0}^{x=1} x^2 dx$ directly from the Riemann sum using the identity $\sum\limits_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}.$

The Riemann sum bridges the gap between single-variable integrals and multiple-variable integrals.

First, let's write down a two variable definite integral; we'll interpret what it means geometrically in a moment: $\int\limits_{x=0}^{x=1} \ \int\limits_{y=0}^{y=1} \left[ x^2+y^2 \right] dx\, dy.$ This looks a lot like the single-variable integral $\int\limits_{x=0}^{x=1} x^2 d x$ we just computed. There, $f(x) = x^2$ measured the height of the point of the curve (parabola) above $x$ on the horizontal axis.

In precisely the same way, $x^2+y^2$ measures the height of a point on a *surface* directly above $(x,y)$ on the plane. In the visualization below, $p =(a,b)$ is an adjustable planar point and $q = (a,b,a^2+b^2)$ is the corresponding point on the surface.

The definite integral
$\int\limits_{x=0}^{x=1} \ \int\limits_{y=0}^{y=1} \left[ x^2+y^2 \right] dx\, dy$
is supposed to be the *volume* contained below this surface and above the square $0 \leq x \leq 1,$ $0 \leq y \leq 1$ in the $xy$-plane.

**Note:** $0 \leq x ,y \leq 1$ means that both $0 \leq x \leq 1$ and $0 \leq y \leq 1$ are true.

Let's write down a Riemann sum approximating the volume contained below the surface $z = f(x,y) = x^2+y^2$ and above the square $0 \leq x \leq 1,$ $0 \leq y \leq 1.$

We use 3D versions of rectangles (called rectangular parallelepipeds) as building blocks for the volume. They have upper right corners sitting at $(x_{i},y_{j} ) = \left( \frac{i}{n} , \frac{j}{m}\right)$ for $i = 1 , \dots , n$ and $j = 1 , \dots , m,$ and the heights are $f(x_{i},y_{j} ).$

If the Riemann sum approximating the volume has the form $\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} \underbrace{f(x_i, y_j)}_{\text{height}} \times \text{base area} ,$ what double sum presented best approximates the total volume below the surface $z = f(x,y) = x^2+y^2$ and above the square $0 \leq x \leq 1, 0 \leq y \leq 1\, ?$

The total volume below the surface $z = x^2 +y^2$ and above the square $0 \leq x \leq 1,$ $0 \leq y \leq 1$ is approximately $\sum_{i=1}^{n}\sum_{j=1}^{m} \left ( \frac{i^2}{n^3 m} +\frac{j^2}{m^3 n}\right).$ If we take the limit as $n , m \to \infty,$ the approximation should approach the exact value.

Use the identity $\sum\limits_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ again to find the total volume.

In single-variable calculus, we learn that the fundamental theorem of calculus can be used to integrate $\int\limits_{x=0}^{x=1} x^2 dx = \int\limits_{x=0}^{x=1} \frac{d}{dx} \left[ \frac{1}{3} x^3 \right] dx = \frac{1}{3} x^3 \Bigg |_{x=0}^{x=1} = \frac{1}{3},$ which matches our Riemann sum result.

Let's see if we can do the same for $\int\limits_{x=0}^{x=1} \ \int\limits_{y=0}^{y=1} \left[ x^2+y^2 \right] dx\, dy.$ This will take some justification later, but assume $\begin{aligned} \int\limits_{x=0}^{x=1} \ \int\limits_{y=0}^{y=1} x^2 d x\, dy & = \ \int\limits_{y=0}^{y=1} \left( \int\limits_{x=0}^{x=1} x^2 d x \right) dy \\ \int\limits_{x=0}^{x=1} \ \int\limits_{y=0}^{y=1} y^2 d x\, dy & = \ \int\limits_{x=0}^{x=1} \left( \int\limits_{y=0}^{y=1} y^2 d y \right) dx,\end{aligned}$ which seems perfectly natural. In words, we integrate in one variable, then the other.

Compute the volume integral using single-variable integration rules.

It seems like the double integral
$\int\limits_{x=a}^{x=b} \ \int\limits_{y=c}^{y=d} f(x,y)\, dx\, dy ,$
which represents the volume below the surface $z = f(x,y) \geq 0$ and above the rectangle $a \leq x \leq b , c \leq y \leq d ,$ can be computed using **nested** single-variable integrals:
$\int\limits_{x=a}^{x=b} \left( \int\limits_{y=c}^{y=d} f(x,y)\, dy \right) dx =\int\limits_{y=c}^{y=d} \left( \int\limits_{x=a}^{x=b} f(x,y)\, dx \right) dy.$
Like partial differentiation, when an integral is performed in one variable, the other is held fixed.

If this way of computing volumes is valid, find the volume below the surface $z= f(x,y) = x e^{y}$ and above the rectangle $0 \leq x \leq 2 , -1 \leq y \leq 1.$

We used the area interpretation of the Riemann sum for a single-variable integral to go one dimension higher into double integration. We used such double integrals to find the volume of two solid regions.

This is just the beginning of a long story we'll take up again in the Multiple Integrals chapter. There, we'll learn how to integrate over more interesting domains, when multiple integrals can be evaluated as nested single-variable integrals, how techniques like $u$-substitution work in higher dimensions, and a few applications of multiple integrals in physics, engineering, and probability.

We hope you enjoyed this brief introduction to the basic ideas underlying multivariable calculus. Our next chapter, Vector Bootcamp, covers the essentials of **vectors**, one of the basic building blocks of our course.

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