Change of variables — general Jacobian

WHAT is going on

Definition Change of variables theorem (2D)

Let $T:(u,v)\mapsto (x,y)$ be a smooth, injective (one-to-one) map with continuous partials on a region, taking region $S$ in the $uv$ -plane onto region $R$ in the $xy$ -plane. Then $\iint_R f(x,y)\,dA = \iint_S f\big(x(u,v),\,y(u,v)\big)\,\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv$ where the Jacobian determinant is $\frac{\partial(x,y)}{\partial(u,v)} = \det\!\begin{pmatrix} x_u & x_v \\ y_u & y_v \end{pmatrix} = x_u\,y_v - x_v\,y_u.$

WHY does the Jacobian appear — derivation from scratch

We derive the 2D scaling factor with no formula memorized.

Take a tiny rectangle in the $uv$ -plane with corner $(u_0,v_0)$ , width $du$ and height $dv$ . Its corners map under $T$ to points in the $xy$ -plane. Using the first-order Taylor (linear) approximation of $T$ near $(u_0,v_0)$ :

Corner stays at $\mathbf{r}(u_0,v_0)$ .
Moving by $du$ in $u$ : displacement $\approx \mathbf{r}_u\,du = (x_u, y_u)\,du$ . Why? The partial derivative is the velocity vector in the $u$ -direction.
Moving by $dv$ in $v$ : displacement $\approx \mathbf{r}_v\,dv = (x_v, y_v)\,dv$ .

So the image is (to first order) a parallelogram spanned by the two edge vectors $\mathbf{a} = (x_u, y_u)\,du, \qquad \mathbf{b} = (x_v, y_v)\,dv.$

The area of a parallelogram spanned by $\mathbf{a},\mathbf{b}$ in 2D is the magnitude of the cross product (the $z$ -component): $\text{Area} = |a_x b_y - a_y b_x| = |x_u\,y_v - y_u\,x_v|\;du\,dv.$

Why this step? In 2D the cross product of $(a_x,a_y,0)$ and $(b_x,b_y,0)$ has only a $z$ -component equal to $a_xb_y - a_yb_x$ ; its magnitude is the area.

Hence $dA = dx\,dy \approx \left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv. \qquad\blacksquare$

Figure — Change of variables — general Jacobian

HOW to use it — recipe

Identify the new variables $u,v$ (often to make $R$ or $f$ simple).
Express $x,y$ in terms of $u,v$ (so you can differentiate easily).
Compute $\dfrac{\partial(x,y)}{\partial(u,v)}$ .
Rewrite $f$ , swap $dA \to |J|\,du\,dv$ , and find the new region $S$ .

Worked Examples

Worked example 2 — A linear change to evaluate

$\iint_R (x+y)\,dA$ $R$ is the parallelogram with vertices $(0,0),(1,1),(2,0),(1,-1)$ . Let $u = x+y,\quad v = x-y \;\Rightarrow\; x = \tfrac{u+v}{2},\ y = \tfrac{u-v}{2}.$ Jacobian: $\frac{\partial(x,y)}{\partial(u,v)} = \det\begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix} = -\tfrac12,\quad |J| = \tfrac12.$ The region becomes $0\le u\le 2,\ -...$ → here $u=x+y$ ranges $0\to2$ and $v=x-y$ ranges $-...$ ; vertices map to $u\in[0,2], v\in[0,2]$ . Then $\iint_R (x+y)\,dA = \int_0^2\!\!\int_0^2 u\cdot\tfrac12\,dv\,du = \tfrac12\int_0^2 2u\,du = \tfrac12\cdot 4 = 2.$ Why this step? $u=x+y$ turns the integrand into a single variable, and the linear map has a constant Jacobian $\tfrac12$ , so it factors right out.

Common Mistakes

Active Recall

Recall Quick self-test (think before peeking)

Why does $dx\,dy = r\,dr\,d\theta$ have an $r$ ? → because a polar cell is a curved wedge whose area grows with radius; Jacobian $=r$ .
What does $\det J$ measure? → local area/volume scaling factor of the linearized map.
Why absolute value? → area is non-negative; sign only encodes orientation flip.

Recall Feynman: explain to a 12-year-old

Imagine drawing a tiny square on a stretchy rubber sheet. When you pull and twist the sheet, that little square becomes a slanted, bigger (or smaller) shape. If you want to add up paint covering the sheet, you must know how much bigger each square got — otherwise you'll think there's more or less paint than there really is. The Jacobian is the number telling you "this square got 3 times bigger" (or half as big). You multiply by it so your total stays honest.

What is the 2D change-of-variables formula?

\iint_R f\,dA = \iint_S f(x(u,v),y(u,v))\,|\partial(x,y)/\partial(u,v)|\,du\,dv

Define the Jacobian determinant

\partial(x,y)/\partial(u,v)

x_u y_v - x_v y_u

, i.e.

\det\begin{pmatrix}x_u&x_v\\y_u&y_v\end{pmatrix}

Why does the Jacobian appear in the integral?

A tiny

uv

-cell maps to a parallelogram whose area is scaled by

|\det J|

; we sum the correct old-area.

Geometric meaning of

\det J

Local signed area/volume scaling factor of the linearized map

T

Why take the absolute value of the Jacobian?

Area must be non-negative; the sign only encodes orientation reversal.

Jacobian for polar coordinates

x=r\cos\theta,y=r\sin\theta

r

, so

dx\,dy=r\,dr\,d\theta

Relation between

\partial(x,y)/\partial(u,v)

and

\partial(u,v)/\partial(x,y)

They are reciprocals: product

=1

(chain rule on inverse maps).

Two edge vectors of the mapped cell?

\mathbf{r}_u\,du=(x_u,y_u)du

and

\mathbf{r}_v\,dv=(x_v,y_v)dv

What requirement on

T

ensures validity?

T

smooth, injective with continuous partials and nonzero Jacobian on the region.

Connections

Polar Coordinates — special case giving $r\,dr\,d\theta$
Spherical and Cylindrical Coordinates — 3D Jacobians ( $\rho^2\sin\phi$ , $r$ )
Determinant as Volume Scaling — linear-algebra root of the idea
Cross Product and Area — used in the derivation
Inverse Function Theorem — guarantees local invertibility when $\det J\neq0$
Double Integrals over General Regions
Chain Rule (Multivariable) — justifies the inverse-Jacobian reciprocal

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab hum double integral mein coordinates change karte hain — jaise $(x,y)$ se $(u,v)$ — tab ek chhota sa rectangle naye coordinates mein purane coordinates mein ek tirchha parallelogram ban jaata hai. Uska area same nahi rehta; woh stretch ya shrink ho jaata hai. Yeh jo scaling factor hai, wahi hai Jacobian determinant ka absolute value. Agar isko bhool gaye to aap area galat ruler se naap rahe ho, aur answer galat aayega.

Jacobian aata kahaan se hai? Microscope se dekho to har smooth map locally linear lagta hai. Tiny cell ki do edges ban jaati hain vectors $\mathbf{r}_u\,du$ aur $\mathbf{r}_v\,dv$ . Inka parallelogram ka area cross product se nikalta hai, aur woh exactly $|x_u y_v - x_v y_u|\,du\,dv$ ho jaata hai. Isiliye $dx\,dy = |J|\,du\,dv$ . Polar mein yeh $r$ ban jaata hai — isliye $dx\,dy = r\,dr\,d\theta$ , woh famous $r$ aise hi aata hai.

Practical tip: agar $x,y$ ko $u,v$ ke terms mein likhna mushkil ho, to ulta $u,v$ ko $x,y$ mein likho, uska Jacobian nikaalo, aur reciprocal le lo (kyunki inverse maps ke determinants reciprocal hote hain). Aur do cheezein hamesha yaad rakhna: (1) absolute value lagao, kyunki area negative nahi hota, aur (2) region ko bhi naye variables mein convert karo, sirf integrand nahi. Yeh do galtiyaan students sabse zyada karte hain.

Change of variables — general Jacobian

WHAT is going on

WHY does the Jacobian appear — derivation from scratch

HOW to use it — recipe

Worked Examples

Common Mistakes

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections