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# 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