Worked examples — Change of variables — general Jacobian

This deep dive is the case gym for the general Jacobian. The parent note showed you the machine; here we feed it every kind of input and watch what happens: nice invertible maps, orientation flips, a Jacobian that hits zero (degenerate!), a limiting-radius sanity check, a real-world word problem, and an exam-style trap.

Before anything, one reminder in plain words so no symbol is unearned:

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The scenario matrix

Every problem this topic throws at you falls into one of these cells. The examples below are each tagged with the cell(s) they cover, so together they cover the whole table.

Cell	What is special	Danger	Covered by
A. Curvilinear standard	polar / smooth nonlinear map	forgetting region change	Ex 1, Ex 6
B. Linear, $J>0$	constant Jacobian, orientation preserved	none, but factor it out	Ex 2
C. Linear, $J<0$	map flips the plane, $\det <0$	dropping the absolute value	Ex 3
D. Inverse shortcut	easier to write $u,v$ in terms of $x,y$	wrong ratio direction	Ex 4
E. Degenerate / $J=0$	Jacobian vanishes somewhere	map not injective there	Ex 5
F. Limiting / sanity	shrink a variable to $0$, check units	area should $\to 0$ correctly	Ex 6
G. Word problem	real quantity (mass/paint)	units and setup	Ex 7
H. Exam twist	they give the wrong-way Jacobian on purpose	reciprocal + $\Vert \cdot \Vert$	Ex 8

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Example 1 — Polar, and where the region actually goes (Cell A)

Steps

1. Map: $x=r\cos \theta ,y=r\sin \theta$. Why this step? The integrand $x^2+y^2$ is $r^2$ — polar makes it one variable, and the round region becomes a rectangle.
2. Jacobian: $J=\det \left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)=r({\cos }^2\theta +{\sin }^2\theta )=r$. Why this step? We need old-area per new-cell; ${\cos }^2+{\sin }^2=1$ collapses it to the clean $r$.
3. New region $S$: the disk $x^2+y^2\le 1$ becomes $0\le r\le 1,0\le \theta \le 2\pi$. Why this step? This is the step people skip. The whole disk is swept once by letting $r$ go $0\to 1$ and $\theta$ go all the way around.
4. Integrate: ${\int }_0^{2\pi }{\int }_0^1{e}^{-r^2}rdrd\theta =2\pi \cdot {\left[-\frac{1}{2}{e}^{-r^2}\right]}_0^1=2\pi \cdot \frac{1}{2}\left(1-e^{-1}\right)=\pi \left(1-e^{-1}\right).$ Why this step? The extra $r$ from the Jacobian is exactly what makes $e^{-r^2}r$ integrable in closed form — without it you'd be stuck.

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Example 2 — Linear map, $J>0$, factor it out (Cell B)

Steps

1. Choose $u=x+y,v=x-y$, so $x=\frac{u+v}{2},y=\frac{u-v}{2}$. Why this step? The edges of the parallelogram lie along $x+y=\text{const}$ and $x-y=\text{const}$ (see figure), so these coordinates straighten $R$ into a square.
2. Jacobian: $\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)=-\frac{1}{2}$, so $|J|=\frac{1}{2}$. Why this step? A linear map has a constant Jacobian — it pulls straight out of the integral. The minus sign just means this map flips orientation (harmless, see Ex 3).
3. Region: the vertices give $u=x+y\in [0,2]$ and $v=x-y\in [0,2]$; $S$ is the square $[0,2]\times [0,2]$. Why this step? Check each vertex: $(0,0)\to (0,0)$, $(1,1)\to (2,0)$, $(2,0)\to (2,2)$, $(1,-1)\to (0,2)$. Full square.
4. Integrate: ${\int }_0^2{\int }_0^{2}u\cdot \frac{1}{2}dvdu=\frac{1}{2}{\int }_0^{2}2udu=\frac{1}{2}\cdot 4=2.$ Why this step? $u=x+y$ is the integrand, so it becomes a single-variable integral times a constant.

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Example 3 — Orientation flip: $J<0$ and the absolute value (Cell C)

Steps

1. Jacobian: $x_u=0,x_v=2,y_u=1,y_v=0$, so $J=x_u{y}_v-x_v{y}_u=0\cdot 0-2\cdot 1=-2$. Why this step? The negative sign is the signal that $T$ reverses orientation — look at the figure: the corner order $u\to v$ runs clockwise in the image where it ran counter-clockwise in the square.
2. Take absolute value: $|J|=2$. Why this step? Area is a length-times-length, always $\ge 0$. If we naively used $-2$ we'd get a negative area — nonsense.
3. Area $={\iint }_{[0,1{]}^2}1\cdot |J|dudv=2\cdot (\text{area of unit square})=2\cdot 1=2.$ Why this step? With $f=1$ the integral just totals the scaled area.

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Example 4 — Inverse-Jacobian shortcut (Cell D)

Steps

1. Compute the easy direction: $\frac{\partial (u,v)}{\partial (x,y)}=\det \left(\begin{array}{cc}{u}_x& u_y\\ v_x& v_y\end{array}\right)=\det \left(\begin{array}{cc}2x& -2y\\ 2y& 2x\end{array}\right)=4x^2+4y^2$. Why this step? $u,v$ are given as functions of $x,y$, so these partials are one-liners — the other direction would need messy square roots.
2. Reciprocate: since $J_T$ and $J_{T^{-1}}$ are inverse matrices, their determinants multiply to $1$ (see Inverse Function Theorem): $\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{4(x^2+y^2)}.$ Why this step? The formula we need for an integral is old-area per new-cell, i.e. $\partial (x,y)/\partial (u,v)$; reciprocal delivers it for free.
3. At $(1,1)$: value $=\frac{1}{4(1+1)}=\frac{1}{8}$. Why this step? Plug in to get a concrete number.

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Example 5 — Degenerate case: the Jacobian is ZERO (Cell E)

Steps

1. Jacobian: $x_u=2u,x_v=0,y_u=0,y_v=1$, so $J=2u\cdot 1-0\cdot 0=2u$. Why this step? We need the local stretch factor as a function of position.
2. Zero locus: $J=0⟺u=0$. Why this step? At $u=0$ the map's $x$-derivative dies — a tiny cell there gets flattened onto a line, area $\to 0$ (look at the collapsed cell in the figure).
3. Injectivity fails: $u=+a$ and $u=-a$ both give $x=a^2$. Why this step? The theorem requires $T$ one-to-one. Here two different $u$'s land on the same $x$, so $T$ folds the plane along $u=0$. You must split into $u\ge 0$ and $u\le 0$ and treat each separately.
4. Consequence: on $u>0$ alone, $|J|=2u$ and $T$ is injective, so $dxdy=2ududv$ is valid there. Why this step? Restricting to a piece where $J\ne 0$ restores all the hypotheses.

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Example 6 — Limiting sanity check with a ring (Cells A, F)

Steps

1. Set up: $\text{Area}={\int }_0^{2\pi }{\int }_a^{b}rdrd\theta$. Why this step? $f=1$, and the Jacobian $r$ is exactly what turns "radial slab" into true area.
2. Evaluate: ${\int }_a^{b}rdr=\frac{1}{2}(b^2-a^2)$, times ${\int }_0^{2\pi }d\theta =2\pi$, giving $\pi (b^2-a^2)$. Why this step? Straight integration; note the $r$ made $rdr$ integrate to $\frac{1}{2}{r}^2$.
3. Limit $a\to b$: $\pi (b^2-a^2)\to \pi (b^2-b^2)=0$. Why this step? Confirms a degenerate (zero-width) ring has zero area — the formula behaves sensibly at the boundary of validity.
4. Sanity at $a=0$: area $=\pi b^2$, the full disk. Why this step? Recovers the known disk area — the formula's other extreme.

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Example 7 — Word problem: mass of a stretched plate (Cell G)

Steps

1. Mass $={\iint }_R\rho dA={\iint }_{[0,1{]}^2}1\cdot |J|dudv$. Why this step? Mass = density integrated over area; with constant $\rho$, mass = $\rho \times (\text{area of }R)$, and area comes from $|J|$.
2. Jacobian: $x_u=1,x_v=1,y_u=1,y_v=-1\Rightarrow J=1\cdot (-1)-1\cdot 1=-2$, so $|J|=2$ (units: m² of plate per m² of $uv$-cell). Why this step? The map doubles areas; the sign again just flags orientation.
3. Mass $={\int }_0^1{\int }_0^{1}2dudv=2\cdot 1=2$ kg. Why this step? Constant integrand over unit square.

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Example 8 — Exam twist: they hand you the wrong-way Jacobian (Cell H)

Steps

1. Check their determinant: $u_x=\frac{1}{y},u_y=-\frac{x}{y^2},v_x=y,v_y=x$. $\frac{\partial (u,v)}{\partial (x,y)}=\det \left(\begin{array}{cc}1/y& -x/y^2\\ y& x\end{array}\right)=\frac{x}{y}-\left(-\frac{x}{y}\right)=\frac{2x}{y}.$ Why this step? The value $\frac{2x}{y}$ is actually correct — but it's the wrong direction for the integral.
2. Error 1 — direction. For $dA=dxdy$ we need $\frac{\partial (x,y)}{\partial (u,v)}$, the reciprocal: $\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{2x/y}=\frac{y}{2x}.$ Why this step? The integral wants old-area ($dxdy$) per new-cell ($dudv$); "new variables go in the denominator" (the JAB rule from the parent).
3. Error 2 — express in $u,v$, and take absolute value. Since $u=x/y$ and $v=xy$, we get $x/y=u$ so $\frac{y}{2x}=\frac{1}{2u}$, and $|J|=\frac{1}{2u}$ (positive since $u>0$). $dA=\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv=\frac{1}{2u}dudv.$ Why this step? The Jacobian must be written in the new variables to actually integrate, and the absolute value keeps area positive.

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Connections

• Change of variables — general Jacobian — the parent machine these examples exercise.
• Polar Coordinates — Ex 1, 6 are its Jacobian $r$ in action.
• Determinant as Volume Scaling — why $\det J$ is the area factor.
• Cross Product and Area — used in Ex 7's verify.
• Inverse Function Theorem — justifies the reciprocal rule (Ex 4, 8) and the $J\ne 0$ requirement (Ex 5).
• Double Integrals over General Regions — where the region-change step (Ex 1) lives.
• Chain Rule (Multivariable) — proves $\det J_T\cdot \det J_{T^{-1}}=1$.