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Change of Variables in Double Integrals

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

Change of Variables in Double Integrals

From single-variable substitution to the Jacobian determinant — the art of re-coordinatizing the plane


1. The Single-Variable Ancestor: Substitution Recalled

1.1 The One-Variable Rule

You already know the substitution rule for a definite integral. If $x = g(u)$ is a differentiable, one-to-one function mapping $[c,d]$ onto $[a,b]$, then

$$ \int_a^b f(x)\,dx \;=\; \int_c^d f\!\bigl(g(u)\bigr)\,\bigl|g'(u)\bigr|\,du. $$

Two things happen simultaneously:

  1. The limits transform — $a$ and $b$ are replaced by their pre-images $c$ and $d$.
  2. The derivative factor is paid — the factor $|g'(u)|$ compensates for the stretching or compressing that $g$ does to the $u$-axis.
The two-variable question: When we re-coordinatize the plane instead of the line, what replaces the derivative factor $|g'(u)|$?
1.2 The Answer: The Jacobian Determinant

The one-dimensional derivative $g'(u)$ measures how much a tiny interval $du$ is stretched into $dx$. In two dimensions, a mapping $T(u,v) = \bigl(x(u,v),\, y(u,v)\bigr)$ stretches tiny rectangles into tiny parallelograms. The factor that measures this local area magnification is the absolute value of the Jacobian determinant:

$$ \left|\frac{\partial(x,y)}{\partial(u,v)}\right| \;=\; \left|\det\begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{pmatrix}\right| \;=\; \left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right|. $$

This is the Jacobian — the machinery we built when we studied local invertibility and area magnification, now installed in the integral formula.

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