When we did u-substitution back in single-variable calculus, we changed the coordinates from x -> u so that we could describe the integral in an easier way. Now, we'll be doing the same thing, but instead of changing the bounds of a range and altering a variable, we will be changing the region of integration along with the variables. Pretty much, this is multivariable equivalent of a u-substitution.

# Jacobian Determinants

# How to solve those problems

First, you have to figure out your **u **and **v**. Usually, it's obvious.

Next, you have to draw your new region. This helps you figure out the bounds. Convert each of the sides of your region to something in terms of u and v.

Once you have your new bounds, you express x and y in terms of u and v, find your Jacobian, and set up your integral. You're done!

# One-to-one

One important thing: when making the transformation, it is important, no required, for it to be one-to-one. One way you could check is if the Jacobian is 0, then it's not one-to-one. However, this requires a very complex proof, so you probably shouldn't use it.

## How to show that something is one-to-one

Here's another problem:

$u = x^2 - y^2, v = 2xy$. We have a problem. If you plug in $(-x,-y)$, you get the same thing! Right away, we know this is not one-to-one

What if we changed it to

$u = x^2 - y^2, v = y$. This is one-to-one. $y = v$, and $x = \pm \sqrt{v^2 - x^2}$. Because there is a $\pm$, then it cannot be one-to-one. If it's one-to-one, it's invertible (we also showed this in the linear algebra blog!), so try to rewrite it as the inverse, like I just did.

Overall, don't worry too much about this, because you will never be wrong. If you try the determinant, and it's 0, then it's not one-to-one. That will always serve as a way to check your work.

David Witten