Part 12 - Change of variables

Introduction

In this part we’ll cover how to perform change of variables for double and triple integrals, as well as the idea and power with polar coordinates.

Change of variable

For functions of one variable:

$$ \int_a^b f(g(x))g’(x)\ dx $$

Let $u = g(x)$ and therefore $du = g’(x)\ dx$.

Then: $$ \int_{g(a)}^{g(b)} f(u)\ du $$

But in two variables, this is a bit tricker, we’ll have to use polar coordinates.

Polar coordinates

Instead of describing a point as $(x, y)$ we can describe it using the magnitude (from the origin) and a angle.

$$ r \geq 0 \newline 0 \leq \theta \leq 2\pi $$

Then: $$ x = r cos(\theta) \newline y = r sin(\theta) $$

When to change to polar coordinates

1. If $f$ looks nicer in polar coordinates

and

2. $D$ looks nice in polar coordinates

Example

Say we have the unit circle: $$ D = \{(x, y) | x^2 + y^2 \leq 1\} $$

In polar coordinates: $$ S = \{(r, \theta) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \} $$

Which is just a rectangle! Which we know how to integrate over!

$$ D = \{(x, y) | 1 \leq x^2 + y^2 \leq 4\} $$

In polar coordinates: $$ S = \{(r, \theta) | 1 \leq r \leq 2, 0 \leq \theta \leq 2\pi \} $$

Or something even more complex: $$ S = \{(r, \theta) | 0 \leq r \leq 1, \theta_1 \leq \theta \leq \theta_2 \} $$

Naive guess

But, how do we change to polar coordinates?

Let’s take a naive guess: $$ \iint_D f(x, y)\ dA = \iint_S f(r cos(\theta), r sin(\theta))\ dA $$

Let’s try!

Given $f(x, y) = 1$ and $D = \{(x,y) | x^2 + y^2 \leq 1\}$.

$$ \iint_D 1\ dA = \text{Area of D} = \pi r^2 = \pi $$

Polar: $$ S = \{(r, \theta) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \} $$

$$ \iint_S 1\ dA = \int_0^{2\pi} \int_0^1 r\ dr\ d\theta = \ldots = \pi $$

So this doesn’t yield the same result.

Definition

$$ \iint_S g(r, \theta)\ dA = \iint_S f(r cos(\theta), r sin(\theta)) \cdot r\ dA $$

Example

Find $\iint_D x^2(x^2 + y^2)^7\ dA$ where $D$ is the region in the upperhalf plane, bounded by the circles, $x^2 + y^2 =1$ and $x^2 + y^2 = 4$.

$$ S = \{(r, \theta) | 1 \leq r \leq 2, 0 \leq \theta \leq \pi \} $$

Which means: $$ \int_0^{\pi} \int_1^2 (r^2 cos^2(\theta) \cdot (r^2)^7) \cdot r\ dA $$

$$ \int_0^{\pi} \int_1^2 r^{17} cos^2(\theta)\ dr\ d\theta = \ldots = \boxed{\pi} $$