Integration of functions

Multi-variable integration

Double Integrals

Definition:

A double integral allows you to integrate a function of two variables over a region in the plane. Given a function $f (x, y)$ and a region $R$ in the $x y$ -plane, the double integral is written as:

\iint_{R} f (x, y) d A

Where $d A$ represents an infinitesimal area element in the plane.

Geometric Interpretation:

The double integral represents the volume under the surface $z = f (x, y)$ over the region $R$ . When $f (x, y) \geq 0$ , the result is the volume trapped between the surface and the plane.

Iterated Integrals:

A double integral can often be computed as two iterated integrals:

\iint_{R} f (x, y) d A = \int_{a}^{b} (\int_{c (x)}^{d (x)} f (x, y) d y) d x

Where $a$ , $b$ , $c (x)$ , and $d (x)$ define the limits of the region $R$ .

Example:

Evaluate the double integral of $f (x, y) = x + y$ over the rectangle $R = [0, 2] \times [0, 3]$ :

\iint_{R} (x + y) d A = \int_{0}^{2} (\int_{0}^{3} (x + y) d y) d x

First, integrate with respect to $y$ :

\int_{0}^{3} (x + y) d y = [x y + \frac{y^{2}}{2}]_{0}^{3} = 3 x + \frac{9}{2}

Then integrate with respect to $x$ :

\int_{0}^{2} (3 x + \frac{9}{2}) d x = {[\frac{3 x^{2}}{2} + \frac{9}{2} x]}_{0}^{2} = (\frac{12}{2} + 9) = 15

So, the value of the double integral is $15$ .

Polar Coordinates and Double Integrals

Definition:

Polar coordinates provide an alternative to Cartesian coordinates for describing locations in the plane. A point in polar coordinates is represented as $(r, θ)$ , where:

$r$ is the distance from the origin to the point.
$θ$ is the angle between the positive $x$ -axis and the line segment connecting the origin to the point.

The conversion between Cartesian coordinates $(x, y)$ and polar coordinates $(r, θ)$ is:

x = r \cos θ, y = r \sin θ

Double Integrals in Polar Coordinates:

To compute a double integral in polar coordinates, use the following formula:

\iint_{R} f (x, y) d A = \iint_{R} f (r \cos θ, r \sin θ) r d r d θ

The extra factor of $r$ comes from the Jacobian determinant when changing variables.

Example:

Find the area of the region inside the circle of radius 2 centered at the origin:

\iint_{R} 1 d A = \int_{0}^{2 π} \int_{0}^{2} r d r d θ

First, integrate with respect to $r$ :

\int_{0}^{2} r d r = {[\frac{r^{2}}{2}]}_{0}^{2} = 2

Now, integrate with respect to $θ$ :

\int_{0}^{2 π} 2 d θ = 2 \times 2 π = 4 π

So, the area of the circle is $4 π$ .

Triple Integrals

Definition:

A triple integral allows you to integrate a function of three variables over a region in three-dimensional space. Given a function $f (x, y, z)$ and a region $D$ in space, the triple integral is written as:

∭_{D} f (x, y, z) d V

Where $d V$ represents an infinitesimal volume element.

Example:

Evaluate the triple integral of $f (x, y, z) = x + y + z$ over the cube $[0, 1] \times [0, 1] \times [0, 1]$ :

∭_{D} (x + y + z) d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (x + y + z) d z d y d x

First, integrate with respect to $z$ :

\int_{0}^{1} (x + y + z) d z = [x z + y z + \frac{z^{2}}{2}]_{0}^{1} = x + y + \frac{1}{2}

Then, integrate with respect to $y$ :

\int_{0}^{1} (x + y + \frac{1}{2}) d y = {[x y + \frac{y^{2}}{2} + \frac{y}{2}]}_{0}^{1} = x + \frac{1}{2} + \frac{1}{2} = x + 1

Finally, integrate with respect to $x$ :

\int_{0}^{1} (x + 1) d x = {[\frac{x^{2}}{2} + x]}_{0}^{1} = \frac{1}{2} + 1 = \frac{3}{2}

So, the value of the triple integral is $\frac{3}{2}$ .

Sequences and Series

Sequences:

A sequence is an ordered list of numbers, typically written as ${a_{n}}$ , where $n$ represents the position in the sequence.

Convergence of a Sequence: A sequence ${a_{n}}$ converges to a limit $L$ if, as $n$ approaches infinity, the terms of the sequence approach $L$ . Formally:

lim_{n \to \infty} a_{n} = L

Divergence: If a sequence does not converge to any finite limit, it is said to diverge.

Series:

A series is the sum of the terms of a sequence. The most common type of series is the infinite series, denoted by:

\sum_{n = 1}^{\infty} a_{n}

Geometric Series: A geometric series has the form:

\sum_{n = 0}^{\infty} a r^{n}

It converges if $| r | < 1$ , and its sum is:

S = \frac{a}{1 - r}

Integral Test:

If $f (x)$ is a continuous, positive, decreasing function, and $a_{n} = f (n)$ , then the series $\sum_{n = 1}^{\infty} a_{n}$ and the improper integral $\int_{1}^{\infty} f (x) d x$ either both converge or both diverge.