4.7.2 Change of variables and the Jacobian

In this lesson

A map from new coordinates $(u, v)$ to $(x, y)$ , and what it does to a grid of small squares
A linear map: every area is multiplied by $\lvert \det A \rvert$ (lesson 3.4.1)
The change-of-variables rule: $dA$ becomes $\lvert \det J \rvert\,du\,dv$
New variables given in terms of $x$ and $y$ , and negative determinants
A non-linear map, its Jacobian matrix of partial derivatives, and an area scale that changes from point to point
Substitution in one variable as the smallest case, and maps that fold

You need double integrals from lesson 4.7.1 and the determinant as an area scale from lesson 3.4.1. The Jacobian uses partial derivatives (lesson 4.1.2), and the last section links back to substitution (lesson 2.6.1).

New coordinates for a region

Some regions are awkward in $x$ and $y$ but simple in other coordinates. A map $(u, v) \mapsto (x, y)$ gives each point of a $uv$ -plane a point of the $xy$ -plane. If it sends a rectangle $S$ onto the region $R$ , an integral over $R$ can be worked out over $S$ instead.

One correction is needed. A small cell of area $du\,dv$ in the $uv$ -plane does not land on a cell of the same area. So each cell must carry the factor by which the map stretches or shrinks area near it.

A linear map

Take the map $x = 3u + v$ , $y = u + 2v$ . In matrix form it is $\mathbf{x} = A\mathbf{u}$ with

A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}

the matrix of lesson 3.4.1. It sends the unit square $S = [0, 1] \times [0, 1]$ to the parallelogram $P$ with corners $(0, 0)$ , $(3, 1)$ , $(4, 3)$ and $(1, 2)$ . Since $\det A = 3 \cdot 2 - 1 \cdot 1 = 5$ , it multiplies every area by 5 (lesson 3.4.1).

The map with matrix $A$ sends a grid of squares to a grid of parallelograms. Every cell's area is multiplied by $\det A = 5$ , so $P$ has area 5.

So a Riemann sum (lesson 4.7.1) over $P$ can use the images of the $uv$ cells. A cell at $(u, v)$ lands near the point $(3u + v, u + 2v)$ and has area $5\,du\,dv$ :

\begin{gathered} \iint_P f(x, y)\,dA \\ = \int_0^1 \int_0^1 f(3u + v,\ u + 2v)\cdot 5\,du\,dv \end{gathered}

For $f(x, y) = x$ this gives $\iint_P x\,dA = 5\int_0^1 \int_0^1 (3u + v)\,du\,dv = 5\left(\frac{3}{2} + \frac{1}{2}\right) = 10$ . Without the factor 5 you would get 2.

The same numbers come from a plain grid in $x$ and $y$ , with no map at all. Cover the box $[0, 4] \times [0, 3]$ with small squares and keep those whose midpoint lies in $P$ , edges included:

squares per unit length	10	100	1000
area of $P$	5.1	5.01	5.001
$\iint_P x\,dA$	10.2	10.02	10.002

The squares cut by the edges make this slow: the error shrinks only in step with the square's width. But both rows settle on 5 and 10.

P

import numpy as np import sympy as sp u, v = sp.symbols("u v") print(sp.integrate(5, (u, 0, 1), (v, 0, 1)), sp.integrate((3*u + v) * 5, (u, 0, 1), (v, 0, 1)), sp.integrate(3*u + v, (u, 0, 1), (v, 0, 1))) for k in [10, 100, 1000]: i = np.arange(4 * k) + 0.5 # x = i / k at the midpoints count = total = 0 for j in np.arange(3 * k) + 0.5: # y = j / k u5 = 2 * i - j # 5k times u v5 = 3 * j - i # 5k times v inside = (u5 >= 0) & (u5 <= 5 * k) & (v5 >= 0) & (v5 <= 5 * k) count += inside.sum() total += i[inside].sum() print(k, round(count / k**2, 4), round(total / k**3, 4))

⌘/Ctrl + Enter runs the cell

Rule: Change of variables, linear map

\begin{gathered} \iint_R f(x, y)\,dA \\ = \iint_S f(x(u, v), y(u, v))\,\lvert \det A \rvert\,du\,dv \end{gathered}

Here $\mathbf{x} = A\mathbf{u}$ sends $S$ onto $R$ . Each $uv$ cell lands on a cell $\lvert \det A \rvert$ times as large, so $dA = \lvert \det A \rvert\,du\,dv$ .

New variables given in terms of $x$ and $y$

Often the region comes as lines in $x$ and $y$ . Take the region $R$ where $0 \le x + y \le 2$ and $0 \le x - y \le 2$ : a square turned through 45°, with corners $(0, 0)$ , $(1, 1)$ , $(2, 0)$ and $(1, -1)$ . In the new variables $u = x + y$ and $v = x - y$ it becomes the square $S = [0, 2] \times [0, 2]$ .

Here the formulas go the other way: they give $(u, v)$ from $(x, y)$ . That map has matrix $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ , with determinant $-2$ . The map back from $(u, v)$ to $(x, y)$ is its inverse, with determinant $\frac{1}{-2} = -\frac{1}{2}$ (lesson 3.4.1).

The minus sign only says the map flips the plane over. The area scale is $\left\lvert -\frac{1}{2} \right\rvert = \frac{1}{2}$ . So the area of $R$ is $\frac{1}{2} \times 4 = 2$ , and

\begin{aligned} \iint_R (x + y)^2\,dA &= \int_0^2 \int_0^2 u^2 \cdot \tfrac{1}{2}\,du\,dv \\ &= \tfrac{1}{2} \cdot \tfrac{8}{3} \cdot 2 = \tfrac{8}{3} \end{aligned}

Rule: Going the other way

\begin{gathered} \text{area scale} = \frac{1}{\lvert \det M \rvert} \end{gathered}

If the new variables are given in terms of the old ones, with matrix $M$ , the map back from $(u, v)$ to $(x, y)$ has determinant $\frac{1}{\det M}$ . Always take the absolute value: an area cannot be negative.

S

import sympy as sp from scipy.integrate import dblquad x, y, u, v = sp.symbols("x y u v") print(sp.Matrix([x + y, x - y]).jacobian([x, y]).det(), sp.Matrix([(u + v) / 2, (u - v) / 2]).jacobian([u, v]).det()) print(sp.integrate(u**2 / 2, (u, 0, 2), (v, 0, 2))) f = lambda y, x: (x + y)**2 left = dblquad(f, 0, 1, lambda x: -x, lambda x: x)[0] right = dblquad(f, 1, 2, lambda x: x - 2, lambda x: 2 - x)[0] print(round(left + right, 10), round(8 / 3, 10))

⌘/Ctrl + Enter runs the cell

A non-linear map

Now take the map $x = u$ , $y = uv$ . Read backwards, $v = \frac{y}{x}$ is the slope of the line from the origin to the point. So the map describes each point by its $x$ and the slope of its ray, a cousin of the distance and angle of lesson 4.7.3.

It sends the unit square $S$ onto the triangle $T$ where $0 \le y \le x \le 1$ . Each line $u =$ constant lands on a vertical segment from $(u, 0)$ to $(u, u)$ , of length $u$ . Each line $v =$ constant lands on the ray $y = vx$ .

The map $x = u$ , $y = uv$ sends the square onto a triangle. Cells near $u = 0$ are squeezed towards the origin; near $u = 1$ they keep most of their area. Each cell's area is multiplied by the $u$ at its centre.

So the area scale changes from point to point. Measure it on a small cell $[u_0, u_0 + h] \times [v_0, v_0 + h]$ .

Its image is a trapezoid with vertical sides of lengths $u_0 h$ and $(u_0 + h)h$ , a distance $h$ apart, so its area is $h^2\left(u_0 + \frac{h}{2}\right)$ . Divide by the cell's area $h^2$ :

side $h$	0.1	0.01	0.001
area ratio at $u_0 = 0.5$	0.55	0.505	0.5005
area ratio at $u_0 = 0.25$	0.3	0.255	0.2505

As the cell shrinks, the ratio settles on $u_0$ : the local area scale at that point.

The Jacobian matrix

A small step $du$ in $u$ moves the point by $\left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right)du$ , the partial derivatives of lesson 4.1.2. A small step $dv$ moves it by $\left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right)dv$ . Close up, the curved grid looks straight, and a small cell lands on the parallelogram with these two sides.

Rule: Jacobian matrix

\begin{gathered} J = \begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[8pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{pmatrix} \\[6pt] \text{local area scale} = \lvert \det J \rvert \end{gathered}

The columns are where small steps in $u$ and in $v$ take the point. By lesson 3.4.1, the parallelogram they span has area $\lvert \det J \rvert\,du\,dv$ .

For a linear map, $J = A$ at every point. Lesson 4.2.2 meets the same matrix in the chain rule.

For $x = u$ , $y = uv$ :

J = \begin{pmatrix} 1 & 0 \\ v & u \end{pmatrix}, \qquad \det J = u

That is the $u_0$ the table found. Along the edge $u = 0$ the determinant is 0: the whole edge lands on the single point $(0, 0)$ . An edge has no area, so this does no harm.

Rule: Change of variables

\begin{gathered} \iint_R f(x, y)\,dx\,dy \\ = \iint_S f(x(u, v), y(u, v))\,\lvert \det J \rvert\,du\,dv \end{gathered}

The map must send $S$ onto $R$ , covering each point of $R$ once. Change three things together: the function (put in $x(u, v)$ and $y(u, v)$ ), the region ( $S$ instead of $R$ ) and the area ( $\lvert \det J \rvert\,du\,dv$ instead of $dx\,dy$ ).

Example. Average the slope $\frac{y}{x}$ over the triangle $T$ . With the map, $\frac{y}{x} = v$ and $dA = u\,du\,dv$ :

\iint_T \frac{y}{x}\,dA = \int_0^1 \int_0^1 v \cdot u\,dv\,du = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

The direct way agrees: $\int_0^1 \int_0^x \frac{y}{x}\,dy\,dx = \int_0^1 \frac{x}{2}\,dx = \frac{1}{4}$ . Without the factor $u$ the answer would be $\frac{1}{2}$ . The area of $T$ is $\int_0^1 \int_0^1 u\,du\,dv = \frac{1}{2}$ , so the average slope, the integral divided by the area, is $\frac{1}{4} \div \frac{1}{2} = \frac{1}{2}$ .

J

import numpy as np import sympy as sp from scipy.integrate import dblquad u, v, x, y = sp.symbols("u v x y") J = sp.Matrix([u, u * v]).jacobian([u, v]) print(J, J.det()) print(sp.integrate(v * u, (v, 0, 1), (u, 0, 1)), sp.integrate(v, (v, 0, 1), (u, 0, 1)), sp.integrate(y / x, (y, 0, x), (x, 0, 1))) print(round(dblquad(lambda y, x: y / x, 0, 1, 0, lambda x: x)[0], 10)) def fan(u, v): return np.array([u, u * v]) def cell_area(p0, p1, p2, p3): t1 = abs(np.linalg.det(np.column_stack([p1 - p0, p2 - p0]))) / 2 t2 = abs(np.linalg.det(np.column_stack([p2 - p0, p3 - p0]))) / 2 return t1 + t2 for u0 in [0.5, 0.25]: v0 = 0.5 ratios = [] for h in [0.1, 0.01, 0.001]: corners = [fan(u0, v0), fan(u0 + h, v0), fan(u0 + h, v0 + h), fan(u0, v0 + h)] ratios.append(round(float(cell_area(*corners) / h**2), 6)) print(u0, ratios)

⌘/Ctrl + Enter runs the cell

Folds, flips and one variable

A flip is fine. The map $(u, v) \mapsto (v, u)$ swaps the axes, a mirror image with $\det J = -1$ . Areas are unchanged, which is why the rule takes $\lvert \det J \rvert$ .

A fold is not. The map $x = u^2$ , $y = v$ sends $[-1, 1] \times [0, 1]$ onto the unit square twice over: $u = 0.5$ and $u = -0.5$ land on the same point. Here $\lvert \det J \rvert = \lvert 2u \rvert$ , and $\int_0^1 \int_{-1}^{1} \lvert 2u \rvert\,du\,dv = 2$ , though the unit square has area 1. A map that covers $R$ twice counts its area twice.

One variable. Substitution (lesson 2.6.1) is the same rule with a $1 \times 1$ Jacobian: $x = g(u)$ gives $dx = g'(u)\,du$ . For $x = 2u + 1$ , which takes $[0, 1]$ to $[1, 3]$ , $\int_1^3 x^2\,dx = \int_0^1 (2u + 1)^2 \cdot 2\,du = \frac{26}{3}$ .

In one variable a falling $g$ needs no absolute value: the new limits run backwards, and swapping them fixes the sign (lesson 2.6.1). A region in the plane has no such order, so the rule uses $\lvert \det J \rvert$ instead.

x = 2u + 1

import sympy as sp u, v, x = sp.symbols("u v x") print(sp.Matrix([v, u]).jacobian([u, v]).det(), sp.Matrix([u**2, v]).jacobian([u, v]).det()) print(sp.integrate(sp.Abs(2 * u), (u, -1, 1), (v, 0, 1))) print(sp.integrate(x**2, (x, 1, 3)), sp.integrate((2*u + 1)**2 * 2, (u, 0, 1)))

⌘/Ctrl + Enter runs the cell

Lesson 4.7.3 applies the rule to polar coordinates, where $\det J = r$ . Case study 4.7.5 uses a linear map to turn independent normal draws into correlated returns. Lesson 5.6.1 uses the rule for functions of random variables.

Common mistakes

Keep the area factor. $\iint_P x\,dA$ is $5 \times 2 = 10$ , not 2: every $du\,dv$ becomes $\lvert \det J \rvert\,du\,dv$ .
Use the right map's determinant. If $u$ and $v$ are given in terms of $x$ and $y$ , the factor is $\frac{1}{\lvert \det M \rvert}$ : $\frac{1}{2}$ for $u = x + y$ , $v = x - y$ , not 2.
Take the absolute value. A determinant of $-\frac{1}{2}$ is an area scale of $\frac{1}{2}$ .
Change the region too. The limits must describe $S$ in $u$ and $v$ , not $R$ .
For a non-linear map, $\det J$ depends on the point. It stays inside the integral: $\iint_T \frac{y}{x}\,dA = \int_0^1\int_0^1 v\,u\,dv\,du$ .
The map must cover $R$ once. $x = u^2$ on $[-1, 1]$ folds, and counts every area twice.

Change of variables

0 of 12 correct

1.
The map $x = 2u - v$ , $y = u + v$ multiplies every area by what factor?
2.
A map has $\det J = -4$ everywhere. A region of area 0.5 in the $uv$ -plane lands on a region of what area?
3.
The matrix $\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ maps the square $[0, 2] \times [0, 2]$ . What is the area of the image?
4.
$P$ is the image of the unit square under $x = 3u + v$ , $y = u + 2v$ . Find $\iint_P y\,dA$
5.
New variables $u = 2x + y$ and $v = x - y$ . By what factor does the map back from $(u, v)$ to $(x, y)$ scale areas?
6.
$R$ is the region where $0 \le 2x + y \le 3$ and $0 \le x - y \le 1$ . Find its area
7.
Find $\det J$ for the map $x = uv$ , $y = v$
8.
Find $\det J$ for the map $x = u^2$ , $y = v^3$ at the point $(u, v) = (1, 2)$
9.
With $x = u$ , $y = uv$ , find $\iint_T x\,dA$ over the triangle $0 \le y \le x \le 1$
10.
The map $x = u$ , $y = uv$ sends $[0, 1] \times [0, 2]$ to the triangle $0 \le y \le 2x \le 2$ . Find its area as $\int_0^2 \int_0^1 u\,du\,dv$
11.
$x = u^2$ , $y = v$ on $[-1, 1] \times [0, 1]$ gives $\iint \lvert \det J \rvert\,du\,dv = 2$ . What is the area of the image, the unit square?
12.
With $x = 2u + 1$ , $\int_1^3 x^2\,dx = \int_0^1 (2u + 1)^2 \cdot k\,du$ . What is $k$ ?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
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Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

(u, v) \mapsto (x, y)

Change of variables and the Jacobian

New coordinates for a region

A linear map

Rule: Change of variables, linear map

New variables given in terms of $x$ and $y$

Rule: Going the other way

A non-linear map

The Jacobian matrix

Rule: Jacobian matrix

Rule: Change of variables

Folds, flips and one variable

Common mistakes

Change of variables

Summary card

Free further reading

New coordinates for a region

A linear map

Rule: Change of variables, linear map

New variables given in terms of xxx and yyy

Rule: Going the other way

A non-linear map

The Jacobian matrix

Rule: Jacobian matrix

Rule: Change of variables

Folds, flips and one variable

Common mistakes

Summary card

Free further reading

New variables given in terms of $x$ and $y$