Change of Variables and the Jacobian

The general substitution rule for double integrals — the Jacobian determinant that makes any change of variables legal.

You've already been changing variables all chapter — polar, cylindrical, and spherical coordinates each swapped x, y, z for friendlier coordinates. This is the general version: any smooth change of variables, with one bookkeeping factor that makes it legal — the Jacobian. Every r and ρ² sin φ you've memorized is really just a Jacobian in disguise; this section reveals the pattern behind all of them.

It's the local stretch factor: when the uv-plane is warped into the xy-plane, a tiny patch of area gets scaled by exactly this amount. It's the same role r played in polar coordinates and ρ² sin φ played in spherical — those are the Jacobians of those specific transformations.

\dfrac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v} - \dfrac{\partial x}{\partial v}\dfrac{\partial y}{\partial u}

The Jacobian (2D)

$u, v$ The new coordinates you transform into.
$\dfrac{\partial(x,y)}{\partial(u,v)}$ The Jacobian — determinant of the matrix of partial derivatives.
$|J|$ Absolute value of the Jacobian — the local area-stretch factor.

Three things happen at once: the integrand gets rewritten in u and v, the region R becomes a simpler region S in the uv-plane, and you multiply by the absolute value of the Jacobian. The absolute value matters — area is positive, so you take |J| even if the determinant itself comes out negative.

\iint_R f(x,y)\, dA = \iint_S f\bigl(x(u,v),\, y(u,v)\bigr)\, \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| \, du\, dv

Change of variables (2D)

$R$ The original region, in the xy-plane.
$S$ The image of R in the uv-plane, under the transformation.

Why bother: the right transformation turns an ugly region — a slanted parallelogram, an awkward curved patch — into a clean rectangle in u, v, and/or turns an ugly integrand into a simple one. You trade a hard region for the cost of computing one determinant.

There's your r. The "extra factor" tacked onto dA in polar coordinates was a Jacobian all along, computed for the specific transformation x = r cos θ, y = r sin θ.

\dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\[6pt] \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r

Sanity check: the polar Jacobian

This is the pattern behind every substitution you've already memorized: the polar Jacobian is r, the spherical Jacobian is ρ² sin φ. Neither is a separate rule — both are this same determinant, computed for their own transformation.

Worked example

\iint_R (x+y)\, dA, \quad R: \text{parallelogram with vertices } (0,0),\ (1,1),\ (2,0),\ (1,-1), \quad u = x+y,\ v = x-y

$u = x+y,\quad v = x-y \;\Longrightarrow\; x = \dfrac{u+v}{2}, \quad y = \dfrac{u-v}{2}$
Solve for x and y in terms of u and v by adding and subtracting the two definitions.
$\dfrac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & -\tfrac{1}{2} \end{vmatrix} = \left(\tfrac{1}{2}\right)\left(-\tfrac{1}{2}\right) - \left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right) = -\dfrac{1}{4}$
Compute the Jacobian determinant from the partial derivatives of x and y with respect to u and v.
$|J| = \dfrac{1}{4}$
Take the absolute value — area is positive, even though the determinant itself came out negative.
$u \in [0,2], \quad v \in [-1,1]$
Run the four vertices of the parallelogram through u = x+y, v = x-y: the slanted region becomes a clean rectangle in the uv-plane. This is the whole payoff of the substitution.
$x + y = u$
The integrand rewrites in the new coordinates about as simply as it can.
$\int_{-1}^{1}\int_0^2 u\cdot\dfrac{1}{4}\,du\,dv = \dfrac{1}{4}\int_{-1}^{1}\left[\dfrac{u^2}{2}\right]_0^2 dv = \dfrac{1}{4}\int_{-1}^{1} 2\,dv = \dfrac{1}{4}\cdot 2 \cdot 2 = 1$
Integrate over the rectangle, carrying the |J| factor along — this is what pays for the simpler region and integrand.

Answer:

1

How to solve one of these

1.
Pick the transformation. Usually chosen to simplify the region (turn a parallelogram or curved patch into a rectangle) or the integrand. Often it's handed to you — the skill is recognizing why it helps.
2.
Solve for x and y in terms of u and v. Or use the given forms directly if $x(u,v)$ and $y(u,v)$ are already provided.
3.
Compute the Jacobian and take its absolute value. Build the $2\times 2$ determinant of partial derivatives, then take $|J|$ — area is always positive.
4.
Transform the region. Find the new bounds in the $uv$ -plane. This is usually the point of the whole exercise: the new region should be simpler than the original.
5.
Rewrite the integrand in u and v. Substitute $x(u,v)$ and $y(u,v)$ into $f(x,y)$ .
6.
Integrate over the new region. Don't forget the $|J|$ factor — it's what makes the substitution legal.

dA = \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| du\, dv

A little bit of area in x, y is the Jacobian's absolute value times a little bit of area in u, v — the local stretch factor.