Part 9 - Double and triple integrals

Introduction

In this part we’ll cover how we integrate functions of several variables, specifically, over a rectangular domain. We’ll later cover general domains.

For now, let’s stick to rectangles and rectangular boxes.

Let’s define our approach.

Double integrals

Say we have a function, $f(x, y)$ with rectangular domain, $R = [a, b] \cdot [c, d]$ .

In mathematical terms: $\{(x, y) | a \leq x \leq b, c \leq y \leq y \}$ .

Let’s now imagine we divide, $[a, b]$ into $n$ subintervals, of width $\Delta x = \dfrac{b - a}{n}$ .

Divide $[c, d]$ into $m$ subintervals, of width $\Delta y = \dfrac{d - c}{m}$ .

Now our rectangle is divided into $nm$ subrectangles, let’s call them, $R_{ij}$ , with area, $\Delta A = \Delta x \Delta y$ .

In each $R_{ij}$ , let’s choose a point denoted as, $(x_{i}^{\*}, y_{j}^{\*})$ .

If we sum up these rectangles:

\sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}) \Delta A

We get something that reminds us of the volume under the graph, taking the limit of these we get:

\lim_{\substack{m \to \infty \newline n \to \infty}} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}) \Delta A

We get the exact volume under the graph (above the rectangle).

Usually, we write this as:

\lim_{\substack{\Delta x \to 0 \newline \Delta y \to 0}} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}) \Delta A

This is the definition of the double integral!

Definition 1 (Double integral)

The double integral of the function, $f$ , on a rectangle, $R$ , is:

\iint_{R} f(x, y)\ dA = \lim_{\substack{\Delta x \to 0 \newline \Delta y \to 0}} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}) \Delta A

Assuming this limit exists, This limit exists for any continuous function.

Notation

We will use this notation:

\iint_{R} f(x, y)\ dA

But we can also write:

\iint_{R} f(x, y)\ dx\ dy

Geometrical sense

$f(x_{i}^{\*}, y_{j}^{\*}) \Delta A =$ volume of rectangular tube

Therefore:

\lim_{\substack{\Delta x \to 0 \newline \Delta y \to 0}} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}) \Delta A \approx \text{ volume under the graph}

Which just means:

\iint_{R} f(x, y)\ dA = \text{ volume under the graph, above } R

General case - functions of three or more variables

The definition is similar, let’s define the triple integral.

Assume that the domain is a rectangular box, $I = [a, b] \cdot [c, d] \cdot [e, f]$

We use the same approach, divide into $n, m, l$ sub-interval, the volume of one of these sub-boxes will be:

f(x_{i}^{\*}, y_{j}^{\*}, z_{k}^{\*}) \Delta V

Summing these:

\sum_{k = 1}^{l} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}, z_{k}^{\*}) \Delta V

Taking the limit:

\lim_{\substack{\Delta x \to 0 \newline \Delta y \to 0 \newline \Delta z \to 0}} \sum_{k = 1}^{l} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}, z_{k}^{\*}) \Delta V

Using integral notation:

\iiint_{I} f(x, y, z)\ dV

How to compute double and triple integrals

Recall how we compute integrals for functions of one variable, specifically recall the fundamental theorem of calculus:

\int_a^b f(x)\ dx = F(b) - F(a)

Where $F(x)$ is the antiderivative/primitive function.

If we can boil it down to an integral of one variable we can use this, this is what we call iterated integrals.

Iterated integrals

Assume we have a function, $f$ , on $R = [a, b] \cdot [c, d]$ .

Fix $y$ , meaning, assume $y$ is a constant.

Therefore:

\int f(x, y)\ dx

is a function that will depend on $y$ , the value we get out from this, is only dependant on $y$ . Therefore:

\int_c^d \left(\int_a^b f(x, y)\ dx \right) dy

We usually just write this as:

\int_c^d \int_a^b f(x, y)\ dx\ dy

This is what we call for iterated integrals. We can also do the other way around!

\int_a^b \int_c^d f(x, y)\ dy\ dx

Notice how our $dx$ and $dy$ are now swapped, this order indicates what order we are integrating.

But does these functions yield the same integral?

Theorem 1 (Fubini’s Theorem)

Let, $f$ , be continuous function on rectangle, $R = [a, b] \cdot [c, d]$ . Then:

\iint_R f(x, y) dA = \int_c^d \int_a^b f(x, y) dx\ dy = \int_a^b \int_c^d f(x, y) dy\ dx

Example 1

Find $\iint_R (x - 3y^2) dA$ , where $R = [0, 2] \cdot [1, 2]$ .

Using Fubini:

\int_1^2 \int_0^2 (x - 3y^2) dx\ dy

\int_1^2 \left(\dfrac{x^2}{2} -3xy^2\right) \bigg\rvert_{x = 0}^{x = 2}\ dy

\int_1^2 (2 - 6y^2) dy

(2y - 2y^3) \bigg\rvert_{y = 1}^{y = 2}

(4 - 16) - (2 - 2) = \boxed{-12}

Let’s do the integral the otherway around as well!

\int_0^2 \int_1^2 (x - 3y^2) dy\ dx

\int_0^2 \left(xy - y^3\right) \bigg\rvert_{y = 1}^{y = 2}\ dx

\int_0^2 (x - 7) dx

(\dfrac{x^2}{2} - 7x) \bigg\rvert_{x = 0}^{x = 2}

(2 - 14) - 0 = \boxed{-12}

Same answer!

Let’s do a triple integral now!

Example 2

Find $\iiint_R xz \cdot e^{xy}\ dV$ on the rectangle, $R = [0, 1] \cdot [0, 1] \cdot [0, 1] = [0, 1]^3$ .

According to Fubini’s Theorem, we can choose any order when computing iterated integrals. If we do a quick analysis, we find that integrating with respect to $z$ first is a good idea.

\int_0^1 \int_0^1 \int_0^1 xz \cdot e^{xy}\ dz\ dx\ dy

\int_0^1 \int_0^1 \dfrac{z^2 x e^{xy}}{2} \bigg\rvert_{z = 0}^{z = 1} dx\ dy

Let’s integrate with respect to $y$ , to make it easier.

\int_0^1 \int_0^1 \dfrac{x e^{xy}}{2} dx\ dy

\int_0^1 \dfrac{e^{xy}}{2} \bigg\rvert_{y = 0}^{y = 1}\ dx

\int_0^1 \dfrac{e^x}{2} - \dfrac{1}{2}\ dx

\dfrac{e^x}{2} - \dfrac{x}{2} \bigg\rvert_{x = 0}^{x = 1}

\left(\dfrac{e^x}{2} - \dfrac{1}{2}\right) - \left(\dfrac{1}{2} - 0\right) = \boxed{\dfrac{e}{2} - 1}

Let’s do one last example.

Example 3

Find $\iint_R y \cdot sin(xy)\ dA$ on the rectangle, $R = [1, 2] \cdot [0, \pi]$ .

Let’s integrate with respect to $x$ , firstly, it’s much easier.

\int_0^{\pi} \int_1^2 y \cdot sin(xy)\ dx\ dy

-\int_0^{\pi} cos(xy) \bigg\rvert_{x = 1}^{x = 2}\ dy

-\int_0^{\pi} cos(2y) - cos(y) \ dy

-\left( \dfrac{sin(2y)}{2} - sin(y)\right) \bigg\rvert_{y = 0}^{y = \pi} = \ldots = \boxed{0}