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Change of Variables

One Dimension: U-Substitution

How do you solve the integral? $$\int_{0}^{5}2xe^{x^2}\, \mathrm{d}x$$ You let $u = x^2$, $du = 2x \mathrm{d}x$ $$\int_{x = 0}^{x = 5}e^{u}\, \mathrm{d}u \to \int_{0}^{25}e^{u}\, \mathrm{d}u = e^{25} - 1$$ if $X$ is our region $[0,5]$, $g(X)$ is our new region $[0,25]$. $$\int_{g(X)}f(u)\, \mathrm{d}u = \int_{X}f(g(x))g'(x)\, \mathrm{d}x$$ Now, how can we turn this logic into one that works in general?

Determinants Represent Scaling Area

$$\to \begin{bmatrix}2& 0 \\ 0 & 3\end{bmatrix} \to $$



The area of the first region is $4 * 4 = 16$. The area of the second region, which is the first region multiplied by the matrix above, is $8 * 12 = 96$. The determinant of the matrix = 6, and 6* the 16 = 96, which begs the question, Does $vol_n T(X) = |det(T)|vol_n(X)$?

Proof

It turns out that it does. The way we show it's true is to verify that it works for the 3 elementary row operations, because if those are true, then you can create any linear transformation from the product of those. We will do a special case of it for now, which is using a rectangle, however the intuition is the same when you generalize for any subset in $\mathbb{R}^n$.

If a rectangle is represented by $[a_1, b_1] \times [a_2, b_2] \times \dots \times [a_n, b_n]$, it's n-dimensional volume is $$\prod_{i = 1}^n (b_i - a_i)$$ $$\text{Case 1: Swapping Rows or } E_I$$ Let's say we swap the i'th row and jth row, the volume stays the same, because we still multiply $(b_k - a_k)$ for all k in $[1,n]$. $$\text{Case 2: Multiply Row by Constant or } E_{II}$$ This becomes the new volume: $(b_1 - a_1) * \dots * (cb_i - ca_i) * \dots * (b_n - a_n) = c\prod_{i = 1}^n (b_i - a_i)$. So, the volume is scaled by c. $$\text{Case 3: Add Multiple of Row to Another Row or } E_{III}$$ Here, each point $x_i$ now equals $x_i + cx_j$. So, the bounds of integration are $(b_1 - a_1) * \dots * (b_i + cx_j - a_i - cx_j) * \dots * (b_n - a_n)$, meaning the volume is maintained. Now, the absolute value of the determinants of the three matrices are 1, c, and 1, respectively. So, when you have a transformation composed of elementary transformations, its determinant is composed of the product of its component determinants. Likewise, the volume changes by the same amount.

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 $$\begin{bmatrix}x \\ y\end{bmatrix} = G\left(\begin{bmatrix}u \\ v\end{bmatrix}\right) = \begin{bmatrix}f(u, v)\\g(u, v)\end{bmatrix}$$ $$\iint_{G(S)} F(x,y)\, dx\, dy = \iint_S F(f(u,v), g(u,v)) \left|\dfrac{\partial (x,y)}{\partial (u,v)}\right|\,du\, dv$$ We can justify this two ways. First, if we consider our argument about areas and determinants, we are integrating over $G(X)$ in the left integral and X on the right, so we have to multiply by the determinant of DG on the right.

Next, if we think of it like u-subs and treat the determinant as a fraction, then you cancel out the $\,du\,dv$.

It's important to note that 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. As a quick recap, here are the 2D and 3D Jacobians. $$\begin{vmatrix}\dfrac{\partial}{\partial u} & \dfrac{\partial}{\partial v}\end{vmatrix} = \begin{vmatrix}\dfrac{\partial x }{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y }{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix}$$ Here is the 3D analog: $$\begin{vmatrix}\dfrac{\partial}{\partial u} & \dfrac{\partial}{\partial v} & \dfrac{\partial}{\partial w}\end{vmatrix} = \begin{vmatrix}\dfrac{\partial x }{\partial u} & \dfrac{\partial x }{\partial v} & \dfrac{\partial x }{\partial w} \\ \dfrac{\partial y }{\partial u} & \dfrac{\partial y }{\partial v} & \dfrac{\partial y }{\partial w} \\ \dfrac{\partial z }{\partial u} & \dfrac{\partial z }{\partial v} & \dfrac{\partial z }{\partial w} \end{vmatrix} $$

Formal Theorem

Let U be an open subset of $\mathbb{R}^n$ and let $g: U \to \mathbb{R}^n$ be a smooth function. Suppose $E$ is a subset of U such that $$1. \text{ E is closed and bounded}$$ This is important because you have to integrate on a compact region, it can't go off to infinity $$2. vol_n(Fr(E)) = 0$$ $$3. \text{ g is injective inside E}$$ This is important because if two points in g collide, then we can't integrate properly. $$4. \forall x \in Int(E), |Dg(x)| \text{ is invertible} $$ Finally, if $f: g(E) \to \mathbb{R}$ is integrable, then so is $(f \circ g)|det[Dg]|: E \to \mathbb{R}$ and $$\int_{g(E)}f = \int_E (f \circ g)|det[Dg]|$$

Special Case: Spherical/Cylindrical

Cylindrical is written like this: $\begin{bmatrix}x \\y\\z\end{bmatrix} = g\left(\begin{bmatrix}r\\ \theta \\ z\end{bmatrix}\right) = \begin{bmatrix}r\cos(\theta)\\ r\sin(\theta) \\ z\end{bmatrix}$ $$\text{The jacobian equals } \begin{bmatrix}\dfrac{\partial}{\partial r} & \dfrac{\partial}{\partial \theta} & \dfrac{\partial}{\partial z}\end{bmatrix} = \begin{bmatrix}\cos(\theta) & -r\sin(\theta) & 0 \\ \sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1\end{bmatrix}$$ The determinant of this equals $r(\cos^2(\theta) + \sin^2(\theta)) = r$ Now, let's do a practice problem: $$\text{Calculate} \iiint\limits_{V} y$$ V is the region below the plan $z = x$ and between the cylinders $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$. We know that the region looks like a washer: a circle with a circle cut out inside. So, $1 \leq r \leq 3$ and $0 \leq \theta \leq 2\pi$. Finally, $0 \leq z \leq r\cos(\theta)$.

Let's quickly go back to our Change of Variables Theorem. We have a function $g(r, \theta, z) \to (x,y,z)$. It's important to remember that by the definition of the Change of Variables Theorem, $g: U \to \mathbb{R}^n$, so it must be of the form (x,y,z) = g(something).
We have a spherical region $E$ and $G(E) = V$. $$\iiint\limits_{G(E)} y = \iiint\limits_{E} r\sin(\theta) * |detJg| = \int_1^2\int_0^{2\pi}\int_0^{r\cos(\theta)} r^2\sin(\theta)\, \mathrm{d}z\,\mathrm{d}\theta\, \mathrm{d}r$$

Spherical is written like this: $(x,y,z) = g(\rho, \theta, \phi)$

How to solve these problems

MathJax TeX Test Page The first thing to is to write the equality $$\int_{G(V)} f = \int_V (f \circ g) |det Jg|$$ Typically, we're given the integral on the left. When we do u-substitution, we get the integral on the right (shown on the top of the page). To illustrate this, let's start with a simple example. $$\text{Evaluate } \iint\limits_{R} x + y \, \mathrm{d}A \text{ where R is the trapezoidal region given by } $$$$(0,0), (5, 0), (\frac{5}{2}, \frac{5}{2}), (\frac{5}{2}, -\frac{5}{2}), x = 2u + v, y = 2u - 3v$$ We see that we have an expression for $x =$ something and $y =$ something. Thus we have a potential function g. $$\begin{bmatrix}x\\y\end{bmatrix} = g\left(\begin{bmatrix}u\\v\end{bmatrix}\right) = \begin{bmatrix}2u+3v\\2u - 3v\end{bmatrix}$$ We take the Jacobian, which is $\begin{bmatrix}2 & 3\\2 & -3\end{bmatrix}$. The determinant is -12 $\neq$ 0, which means it's both invertible, therefore it's injective. Finally, the determinant is never 0 inside the region, so we can take its inverse.

Now, we write the shell of the problem, which is $$\iint_R x + y \, \mathrm{d}A = \iint_{other} \left((2u + 3v) + (2u - 3v)\right)*12\, \mathrm{d}u\mathrm{d}v$$ We have $R$ on the left. We want the left to be G(something). We just let $other = G^{-1}(R)$, that way, taking G(other) = R. $$\iint_R x + y \, \mathrm{d}A = \iint_{G^{-1}(R)} \left((2u + 3v) + (2u - 3v)\right)*12\, \mathrm{d}u\mathrm{d}v$$ The way we solve this is by plugging in the sides of the region. $$y = x \to 2u + 3v = 2u - 3v \to v = 0$$ $$y = x- 5 \to 2u + 3v = 2u - 3v - 5 \to v = \frac{5}{6}$$ $$y = -x \to 2u + 3v = -2u + 3v \to u = 0$$ $$y = -x + 5 \to 2u + 3v = -2u + 3v + 5 \to u = \frac{5}{4}$$ We now have our $G^{-1}(R)$. The reason it's inverted is $G: (u,v) \to (x,y)$, and we're going the other way around. $$\iint_R x + y \, \mathrm{d}A = \int_0^{\frac{5}{6}}\int_0^{\frac{5}{4}} \left(4u\right)*12\, \mathrm{d}u\mathrm{d}v$$ $$ = 10\int_0^{\frac{5}{4}} \left(4u\right)\, \mathrm{d}u = 20(5/4)^2 = 31.25$$

What if we don’t know g?

MathJax TeX Test Page Let's do this problem $$\iint (x+y)e^{xy}\,\mathrm{d}A$$ A is the region enclosed by $\begin{cases}x-y = 1\\x-y = 4\\xy = 1\\xy = 2\end{cases}$ So, we say $u = x-y, v = xy$. However, we do not have an expression for $x$ or $y$. We have a $g^{-1}$. Its Jacobian is $$\begin{bmatrix}1 & -1\\y & x\end{bmatrix}$$ The determinant is $x + y$ which is always positive, because x and y are always positive in this region.

We start by writing the region on the left side $$\iint_A (x+y)e^{xy}\,\mathrm{d}A = \iint_{G^{-1}(A)} (x+y)e^{xy}| det Jg|\,\mathrm{d}u\,\mathrm{d}v$$ We don't know what $x+y$ equals, but we do know $|det Jg|$, it's just $\dfrac{1}{|det Jg^{-1}|} = \dfrac{1}{x + y}$ $$\iint_A (x+y)e^{xy}\,\mathrm{d}A = \iint_{G^{-1}(A)} (x+y)e^{xy}\dfrac{1}{x+y}\,\mathrm{d}u\,\mathrm{d}v$$ $$\iint_A (x+y)e^{xy}\,\mathrm{d}A = \iint_{G^{-1}(A)} e^{xy}\,\mathrm{d}u\,\mathrm{d}v$$ We know $g(xy) = v$, because $g^{-1}(v) = xy$. Now, we can solve the integral $$\iint_A (x+y)e^{xy}\,\mathrm{d}A = \int_{1}^{4}\int_{1}^{2} e^{v}\,\mathrm{d}v\,\mathrm{d}u = 3(e^2 - e) = 14.012$$ We get this same answer when we actually solve the integral, which is very computationally intensive and it requires splitting up into 3 smaller integrals.
David Witten

Surface Integrals

Independence of Path