## Feb 1 Change of Variables

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

MathJax TeX Test Page The way we write change of variable integrals is like u-substitution. This is what u-sub looks like $$\int f(g(x)) g'(x)\, dx = \int f(u)\, du , u = g(x)$$ Now, we say $$x = f(u, v), y = g(u, v)$$ $$\iint_R F(x,y)\, dx\, dy = \iint_S F(f(u,v), g(u,v)) \left|\dfrac{\partial (x,y)}{\partial (u,v)}\right|\,du\, dv$$ A hand-wavy justification of this is if we treat the partial as a fraction, then you cancel out the $\,du\,dv$. Also, it's the absolute value of the determinant. Now, we get to how we write it. Note: when we write matrices like these, we mean the determinant. If you don't know how to calculate a determinant, go to my post about it. $$\begin{vmatrix}\dfrac{\partial}{\partial u} & \dfrac{\partial}{\partial v}\end{vmatrix} = \begin{vmatrix}\dfrac{\partial x }{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y }{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix}$$ Here is the 3D analog: $$\begin{vmatrix}\dfrac{\partial}{\partial u} & \dfrac{\partial}{\partial v} & \dfrac{\partial}{\partial w}\end{vmatrix} = \begin{vmatrix}\dfrac{\partial x }{\partial u} & \dfrac{\partial x }{\partial v} & \dfrac{\partial x }{\partial w} \\ \dfrac{\partial y }{\partial u} & \dfrac{\partial y }{\partial v} & \dfrac{\partial y }{\partial w} \\ \dfrac{\partial z }{\partial u} & \dfrac{\partial z }{\partial v} & \dfrac{\partial z }{\partial w} \end{vmatrix}$$

# How to solve those problems

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

MathJax TeX Test Page $\iint (x-y)^2 \cos^2(x + y) \,dx\,dy$ has $u = x-y$, and $v = x + y$.

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

MathJax TeX Test Page To know how to prove something as one-to-one, it's important to understand what it means. Basically, for every element in the range, there is at most one element in the domain. For example, $y = x^2$ is not one-to-one, $y = 4$ corresponds to two x values: 2 and -2. This is from a Jacobian problem: $u = x^3, v = x + y$. $T(x,y) = (u, v)$. For every u, there is one x value, because $u = x^3$ is one-to-one. Now, v is kind of tricky. Because x is one-to-one, you can just write it as $v = \sqrt[3]{x}+ y$. Given a point (u,v), there is no ambiguity for what the x or y are.
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.