General Changes of Variables and Jacobians

A derivative matrix scales small areas

For a differentiable map $(x,y)=T(u,v)$ , the Jacobian determinant is

J=\det\frac{\partial(x,y)}{\partial(u,v)}=x_uy_v-x_vy_u.

Small parameter rectangles become approximate parallelograms with area scale $|J|$ . Under a suitable one-to-one smooth change of variables with nonzero Jacobian in the interior,

\iint_R f(x,y)dA=\iint_S f(T(u,v))|J(u,v)|du dv.

Standard extensions allow some boundary degeneracies. If a map covers the same area repeatedly, the ordinary one-to-one formula must be adjusted; an absolute determinant does not correct multiplicity.

Visual guide

VISUAL GUIDEA square maps to a parallelogram of twice the area

Under x = u + v, y = u − v, the unit parameter square becomes the shaded diamond. The Jacobian determinant is −2: magnitude 2 scales area, while the negative sign records orientation reversal.

Worked example: turn a parallelogram into a square

Let $x=u+v$ , $y=u-v$ with $0\le u,v\le1$ . The unit square maps to vertices $(0,0),(1,1),(2,0),(1,-1)$ . The Jacobian is $-2$ , so geometric area is $\int_0^1\int_0^12du dv=2$ .

To integrate $x+y$ , substitute $x+y=2u$ . The result is $\int_0^1\int_0^1(2u)2dvdu=2$ . The negative determinant records orientation reversal; the area integral uses magnitude.

Worked example: an ellipse

Map the unit disk by $x=au$ , $y=bv$ with $a,b>0$ . The image is $x^2/a^2+y^2/b^2\le1$ and $|J|=ab$ . Its area is therefore $ab\pi$ .

Combining this scaling with polar coordinates gives $x=ar\cos\theta$ , $y=br\sin\theta$ and total Jacobian $abr$ . Jacobians multiply under composition, matching the chain rule for derivative matrices. This is often simpler than computing a large determinant from scratch.

If the available formula gives the inverse transformation, its determinant is reciprocal at corresponding regular points. Be explicit about which direction is being differentiated before inserting a factor into an integral.

Practice

Find the Jacobian for $x=2u,y=3v$ .
Find the area of the ellipse $x^2/4+y^2/9\le1$ .
Why does a determinant of zero prevent a regular local area conversion?

Show worked solutions

$J=6$ .
Semiaxes are $2,3$ , so the area is $6\pi$ .
The derivative collapses at least one direction, so it is not locally invertible by the usual inverse-function theorem.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.