Change of variables — general Jacobian: Visual walkthrough

We use two flat pages side by side:

the input page, where our coordinates are called $u$ (horizontal) and $v$ (vertical);
the output page, where coordinates are called $x$ (horizontal) and $y$ (vertical).

A rule $T$ takes each point $(u, v)$ on the input page and drops it somewhere $(x, y)$ on the output page. Read $T$ as a machine: feed in $(u, v)$ , out comes $(x, y)$ .

Step 1 — Two pages and one machine between them

WHAT. We draw the input page (left) and the output page (right). We mark one point $(u_{0}, v_{0})$ on the input page and follow it through the machine $T$ to its landing spot on the output page.

WHY. Before anything moves or stretches, we must agree what "the machine" even means. Everything later is just: how do points near $(u_{0}, v_{0})$ move compared to $(u_{0}, v_{0})$ itself? So we pin down that home base first.

PICTURE. Look at the magenta dot on the left and the orange dot on the right — same point, before and after $T$ .

Figure — Change of variables — general Jacobian

Step 2 — Draw the tiny cell we care about

WHAT. On the input page we draw a tiny rectangle: its bottom-left corner sits at $(u_{0}, v_{0})$ , it has width $d u$ (a tiny step in the $u$ -direction) and height $d v$ (a tiny step in the $v$ -direction).

WHY. An integral is a sum over tiny cells. If we understand how one tiny cell's area changes, we understand the whole integral — because the map looks the same everywhere at small enough scale. We pick a rectangle because rectangles tile the plane perfectly, so summing them recovers the region.

PICTURE. The violet rectangle has four labelled corners. Its input area is dead simple: $d u \times d v$ .

Step 3 — Where do the corners land? (velocity vectors)

WHAT. We ask: when I nudge the input by $d u$ in the $u$ -direction, how far and in what direction does the output move? The answer is the partial derivative vector $r_{u}$ , times $d u$ .

WHY THIS TOOL — the partial derivative. We need a way to say "how fast does the output slide when I wiggle only $u$ ." That is exactly what a partial derivative measures: the rate of change of a function when one input moves and the others are frozen. We use $\frac{\partial x}{\partial u}$ (written $x_{u}$ ) rather than an ordinary derivative precisely because there are two inputs and we're freezing $v$ .

PICTURE. From the orange landing dot, two arrows sprout: the magenta arrow is the image of the $d u$ -step, the violet arrow is the image of the $d v$ -step.

Step 4 — The cell becomes a parallelogram

WHAT. The two mapped edge-arrows $a$ and $b$ span a shape on the output page. Since they start from the same corner and are two straight edges, the image of our tiny rectangle is (to first order) a parallelogram — the fourth corner is just $a + b$ away.

WHY. The original was a rectangle whose two edges were " $d u$ across" and " $d v$ up." A linear map sends straight edges to straight edges and keeps parallel edges parallel — so a rectangle can only become a parallelogram (possibly rotated, stretched, sheared). It is generally not a rectangle: the two arrows need not be perpendicular anymore. That tilt is exactly why the naive "new width × new height" would be wrong.

PICTURE. The violet input rectangle and the filled parallelogram it maps to, with the fourth corner reached by adding $a$ and $b$ tip-to-tail.

Step 5 — The area of a parallelogram (why the cross product)

WHAT. We compute the area of the parallelogram spanned by $a = (a_{x}, a_{y})$ and $b = (b_{x}, b_{y})$ . The answer is $∣ a_{x} b_{y} - a_{y} b_{x} ∣$ .

WHY THIS TOOL — the cross product. Area of a slanted parallelogram is $(base) \times (true height)$ , and the true height involves the sine of the angle between the edges. The cross product packages exactly that: $∣ a \times b ∣ = ∣ a ∣ ∣ b ∣ sin ϑ$ is the area, and it automatically vanishes when the edges are parallel (zero area). In 2D, embedding the arrows in 3D as $(a_{x}, a_{y}, 0)$ and $(b_{x}, b_{y}, 0)$ , the cross product points straight out of the page with just a $z$ -component: $a \times b = (0, 0, a_{x} b_{y} - a_{y} b_{x}) .$

PICTURE. The parallelogram with base $∣ a ∣$ (magenta), the perpendicular height dropped from the tip of $b$ (dashed), and the angle $ϑ$ between the edges marked.

Step 6 — Substitute the arrows: the Jacobian is born

WHAT. Plug $a = (x_{u}, y_{u}) d u$ and $b = (x_{v}, y_{v}) d v$ into the parallelogram-area formula.

WHY. We finally connect geometry (area of the mapped cell) to calculus (the partial derivatives of $T$ ). This substitution is the whole climax.

PICTURE. The same parallelogram, now with each arrow labelled by its component formula, and the resulting area expression written across it.

Step 7 — The sign, and the degenerate (zero-area) case

WHAT. We examine what the sign of $x_{u} y_{v} - x_{v} y_{u}$ means before we throw it away with $∣ \cdot ∣$ , and what happens when it is zero.

WHY. The contract: cover every case. A positive Jacobian, a negative one, and a zero one are three genuinely different geometric situations, and the reader must recognize each.

PICTURE. Three little cells side by side: orientation preserved (positive), flipped like a mirror (negative), collapsed to a line segment (zero).

The one-picture summary

Everything above compressed into a single diagram: input rectangle $d u d v$ → linear approximation via velocity arrows $r_{u} d u, r_{v} d v$ → parallelogram → its area is $∣ det J ∣ d u d v$ .

Recall Feynman retelling — say it to a friend

Picture graph paper drawn on a stretchy sheet. Pick one tiny square: it's $d u$ wide and $d v$ tall, so its area is $d u d v$ — easy. Now pull and twist the sheet. Track the square's bottom-left corner as home base. When you nudge home base rightward by $d u$ , the corner next to it slides to some new spot — call that arrow $a$ . Nudge upward by $d v$ instead, and the other neighbour slides along arrow $b$ . Those two arrows are the sides of the new shape, and since straight edges stay straight, the shape is a slanted parallelogram. Its area isn't width-times-height (the sides are tilted), it's $∣ a_{x} b_{y} - a_{y} b_{x} ∣$ — the cross-product rule, which quietly subtracts away the slant and gives zero when the sides line up. Feed in what $a$ and $b$ actually are, and that expression is $∣ x_{u} y_{v} - x_{v} y_{u} ∣ d u d v$ : the tiny square's area got multiplied by the Jacobian. A negative Jacobian just means the sheet got mirror-flipped, so we use its size only. And if the Jacobian is zero, the square got crushed to a line — the map isn't reversible there. Multiply by this stretch factor for every square, add them up, and your total (of paint, mass, probability — whatever $f$ is) stays honest.

Recall Quick self-test

Why is the mapped cell a parallelogram and not a rectangle? ::: The two edge-arrows $r_{u} d u$ and $r_{v} d v$ need not be perpendicular; a linear map keeps edges straight and parallel but tilts them. Where does the subtraction $x_{u} y_{v} - x_{v} y_{u}$ come from? ::: From the cross-product area formula $a_{x} b_{y} - a_{y} b_{x}$ ; the minus sign cancels the slant and gives $0$ for parallel edges. Why $∣ det J ∣$ and not $det J$ ? ::: Area is non-negative; a negative determinant only records an orientation flip (mirror image). What does $det J = 0$ mean geometrically? ::: The mapped cell collapsed to a segment (edges parallel); $T$ is not locally invertible there.

Parent: Change of variables — general Jacobian