Change of variables and Jacobians

Transform both the region and the area element

A coordinate map $T(u,v)=(x(u,v),y(u,v))$ can turn a complicated region into a simple one. It also stretches small areas. At a regular point, the linear approximation $DT$ sends a small coordinate rectangle to an approximate parallelogram of area $|\det DT|\,du\,dv$. The determinant's absolute value measures unsigned area, while its sign records orientation.

Definition

Jacobian determinant

For a $C^1$ planar map $T(u,v)=(x(u,v),y(u,v))$, its Jacobian determinant is

$$J_T(u,v)=\frac{\partial(x,y)}{\partial(u,v)} =\det\begin{pmatrix}x_u&x_v\\y_u&y_v\end{pmatrix}.$$

Theorem

Change of variables in two dimensions

Let $S\subset\mathbb R^2$ be bounded and Jordan measurable, with compact closure $\overline S$ contained in an open set $U$. Suppose $T:U\to V$ is a $C^1$ diffeomorphism onto an open set $V$, let $D=T(S)$, and let $f$ be continuous on $T(\overline S)$. Then

$$\iint_D f(x,y)\,dx\,dy =\iint_S f(T(u,v))\,|J_T(u,v)|\,du\,dv.$$

For a bounded set, Jordan measurable means that its boundary has area zero. A $C^1$ diffeomorphism is a bijective $C^1$ map with a $C^1$ inverse; consequently its Jacobian is nonzero. These are sufficient hypotheses; a map covering a positive-area part of $D$ multiple times cannot be used in this formula without accounting for multiplicity.

Proof

Geometric reason for the determinant

For small increments $\Delta u,\Delta v$, differentiability approximates the image of a coordinate cell by the parallelogram generated by $T_u\Delta u$ and $T_v\Delta v$. Its area is $|\det(T_u,T_v)|\Delta u\Delta v$. Summing transformed sample values and taking limits gives the formula under hypotheses ensuring that the local approximations and partition errors can be controlled. This is the idea, not a full proof for arbitrary regions.

Polar coordinates

Set $x=r\cos\theta$, $y=r\sin\theta$, with $r\ge0$. Then

$$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| =\left|\det\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix}\right|=r.$$

Thus $dA=r\,dr\,d\theta$. For a disk of radius $R>0$ and an integrand continuous on the closed disk, first restrict to $\varepsilon\le r\le R$ and $\delta\le\theta\le2\pi-\delta$, with $0<\varepsilon<R$ and $0<\delta<\pi$. The map is a diffeomorphism on an open neighbourhood of this rectangle. Apply the theorem there, then let $\varepsilon,\delta\to0$: the omitted disk and angular wedge have area tending to zero, and continuity on the closed disk makes the integrand bounded; the corresponding parameter integrand is also bounded, so the omitted integrals tend to zero. This justifies the disk formula despite the origin and seam, rather than treating them as an automatic exception. An angular interval of length $4\pi$ would count almost every point twice.

Worked example

A disk and a radial integrand

For the disk $D:x^2+y^2\le4$,

$$\iint_D(x^2+y^2)\,dA=\int_0^{2\pi}\int_0^2r^2\,r\,dr\,d\theta=8\pi.$$

The first $r^2$ is the transformed integrand; the additional $r$ is the Jacobian. Forgetting it changes the answer and the dimension of the area element.

Worked example

A linear transformation

Let $D$ be the parallelogram $0\le x+y\le2$, $0\le x-y\le1$, and evaluate $\iint_D(x+y)\,dA$. Set $u=x+y$, $v=x-y$, so $x=(u+v)/2$, $y=(u-v)/2$. The new region is $0\le u\le2$, $0\le v\le1$, and $|\partial(x,y)/\partial(u,v)|=1/2$. Therefore

$$\iint_D(x+y)\,dA=\int_0^2\int_0^1u\cdot\frac12\,dv\,du=1.$$

Note that $\partial(u,v)/\partial(x,y)=-2$ is the inverse Jacobian; its absolute reciprocal is the factor needed here.

Comparison and checks

Rectangular coordinates are often best for regions with simple vertical or horizontal bounds. Polar coordinates suit disks, sectors, and radial functions; a linear substitution suits parallelograms and linear combinations. Always (1) describe the new region, (2) express the integrand in new variables, (3) compute the correct-direction Jacobian, (4) check one-to-one coverage, and (5) verify boundary and sign conventions. Negative Jacobian does not make an ordinary scalar area integral negative.

Practice

1. Compute the area of $x^2+y^2\le9$ by polar integration.
2. Evaluate $\iint_D e^{x^2+y^2}\,dA$ for $D:x^2+y^2\le1$.
3. Let $x=2u$, $y=3v$. Find the image of $[0,1]^2$ and compute its area using a Jacobian.
4. Explain why $0\le r\le1$, $0\le\theta\le4\pi$ is not a one-to-one polar parametrization of the unit disk up to boundary exceptions.

Full solutions

1. $\int_0^{2\pi}\int_0^3r\,dr\,d\theta=9\pi$.
2. $\int_0^{2\pi}\int_0^1e^{r^2}r\,dr\,d\theta=\pi(e-1)$.
3. The image is $[0,2]\times[0,3]$, and the Jacobian is $6$. Its area is $\int_0^1\int_0^1 6\,du\,dv=6$.
4. Every nonzero interior point has two representatives with angles separated by $2\pi$, away from a negligible seam. The integral over this parameter range counts its area twice.

Synthesis

A substitution changes coordinates, the integrand, the domain, and the measure together. The Jacobian accounts for local area scaling, while injectivity or explicit multiplicity control prevents overcounting.