The Change of Variable Formula

Suppose we have a coordinate system:

Observe that this is kind of like basis vectors, but is like , and is like .

If the region in is:

we wish to transform the region $R$ into the $(x, y)$ coordinate system.

(note that the arrows are just for visualization, sort of like a parameterization)

Fact: Boundaries are mapped to boundaries

For :

• and
• We have

using , we have and , so .

For :

• and
• We have

using , we have and , so .

For :

• and
• We have

using , we have and , so .

For :

• and
• We have

using , we have and , so

Given a function that maps , then

where

which is known as the Jacobian of the transformation.

(see determinant, Partial Derivatives).

Recall that the determinant represents the change in area of a transformation. Also recall from change of basis that changing basis is simply a matrix, where the determinant is the change in area of said matrix, and the change-of-basis matrix can be interpreted as a linear transformation.

So, the Jacobian represents the change in area of the tangent vectors, and we just put it together with the area element so we're integrating with the correct infinitesimal. This is because our change-of-variable transformation "distorts" the area, so we just have to factor that in.

In other words, .

Another way to look at this, is when doing Method of Substitution, we differentiated the substitution. The derivative is the single-variable version of the Jacobian.

Recall that a non-zero determinant means a linear transformation is invertible. The same thing applies, where should not equal , so our change of variable is invertible.

The transformation to Polar Coordinates:

So

As we've seen before with Double Integrals in Polar Coordinates.

(see Pythagorean identities).

Evaluate where is the region bound by and and .

solution
Using our example from before (#^d3d1de),

Our integral is very ugly: $\displaystyle \int_{0}^{2} \int_{y^2/4 - 1}^{1 - y^2/4} y \, dx \, dy$

But we can transform to and . We also know the bounds already (check referenced example).

Our Jacobian:

So we have:

Evaluate , where is the interior of the ellipse

solution
We don't know how to evaluate integrals over ellipses, but we do know how to evaluate an integral over a circle.

We will map the ellipse onto a unit circle. Let and . Then we have:

Our transformation shrinks the domain considerably.

left=-6; right=6;
bottom=-4; top=4;
---
9x^2 + 4y^2 = 36
x^2 + y^2 = 1

The Jacobian is:

This compensates for the shrinking of the domain during integration. Now we have

The domain is circular, and now we can integrate in polar coordinates. Let and , and then

Evaluate where is the region:

using and .

solution
Adding the two equations gives . Subtracting the two equations gives . Changing variables to and , our integral becomes:

Our Jacobian:

What does our new shape look like? We can use the strategy from #^d3d1de to map boundaries to boundaries:

which gives us the new shape:

First, we integrate "horizontally" (), which is bounded between the lines and , so the bounds are .
Next, we integrate "vertically" (), which is bounded between and , so the bounds are .

Now we have