Part 12 - Change of variables

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.

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 trickier, we’ll have to use polar coordinates.

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

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

Then:

\begin{align*} x & = r \cos(\theta) \newline y & = r \sin(\theta) \end{align*}

and

Example 1

Say we have the unit circle:

D = \{(x, y) \mid x^2 + y^2 \leq 1\}

In polar coordinates:

S = \{(r, \theta) \mid 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) \mid 1 \leq x^2 + y^2 \leq 4\}

In polar coordinates:

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

Or something even more complex:

S = \{(r, \theta) \mid 0 \leq r \leq 1, \theta_1 \leq \theta \leq \theta_2 \}

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

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

Polar:

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

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

So this doesn’t yield the same result.

Definition 1 (Double integral in polar coordinates)

\iint_D f(x, y) \, dA = \iint_S f(r \cos(\theta), r \sin(\theta)) \cdot r \, dr \, d\theta

Example 2

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

S = \{(r, \theta) \mid 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 \, dr \, d\theta

\int_0^{\pi} \int_1^2 r^{17} \cos^2(\theta) \, dr \, d\theta = \ldots = \boxed{\dfrac{(2^{18} - 1)\pi}{36}}