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
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?
Fubini’s Theorem
Fubini’s Theorem says that:
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
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
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
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. $$ \int_0^{\pi} \int_1^2 y \cdot sin(xy)\ dx\ dy $$
$$ -\int_0^{\pi} cos(xy) \bigg\rvert_{x = 1}^{x = 2}\ dy $$
$$ -\int_0^{\pi} cos(2y) - cos(y) \ dy $$
$$ -\left( \dfrac{sin(2y)}{2} - sin(y)\right) \bigg\rvert_{y = 0}^{y = \pi} = \ldots = \boxed{0} $$