Change of Variables & the Jacobian

Switching to polar coordinates introduced a mysterious extra r in dA = r\, dr\, d\theta. That r was no coincidence — it is one instance of a single master formula. For any smooth substitution (u, v) \mapsto (x, y),

\iint_R f(x, y)\, dx\, dy = \iint_S f\big(x(u,v),\, y(u,v)\big)\; |J|\; du\, dv,

where R is the region in the xy-plane, S is its preimage in the uv-plane, and |J| is the absolute value of the Jacobian determinant. This is the multivariable cousin of u-substitution, where the one-dimensional stretch factor \frac{dx}{du} is replaced by a determinant.

The Jacobian: the local area-scaling factor

The Jacobian of the map (u,v) \mapsto (x,y) is the determinant of the matrix of partial derivatives:

J = \frac{\partial(x, y)}{\partial(u, v)} = \det\!\begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = x_u\, y_v - x_v\, y_u.

Why it is exactly the area factor. Take a tiny square in the uv-plane with corner (u, v) and sides du, dv.

Step 1 — see where the two edges go. The edge along u (length du) maps, to first order, to the vector (x_u, y_u)\, du; the edge along v maps to (x_v, y_v)\, dv. The image of the little square is the parallelogram spanned by these two vectors.

Step 2 — area of a parallelogram is a determinant. The area spanned by vectors (a, c) and (b, d) is |ad - bc|. With (a,c) = (x_u, y_u)\,du and (b,d) = (x_v, y_v)\,dv:

\text{image area} = |x_u\, y_v - x_v\, y_u|\; du\, dv = |J|\; du\, dv.

So |J| is precisely the factor by which the map stretches area near a point. A |J| > 1 blows patches up, |J| < 1 shrinks them, and that is why it must appear to keep the integral honest.

Worked example 1 — polar, recovering the r

Let us check the formula against a case we already trust. The polar map is x = r\cos\theta, y = r\sin\theta (so here (u, v) = (r, \theta)).

Step 1 — the four partial derivatives.

x_r = \cos\theta, \quad x_\theta = -r\sin\theta, \quad y_r = \sin\theta, \quad y_\theta = r\cos\theta.

Step 2 — assemble and take the determinant.

J = \det\!\begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{bmatrix} = (\cos\theta)(r\cos\theta) - (-r\sin\theta)(\sin\theta).

Step 3 — simplify with \cos^2 + \sin^2 = 1.

J = r\cos^2\theta + r\sin^2\theta = r\,(\cos^2\theta + \sin^2\theta) = r.

Hence dx\, dy = |J|\, dr\, d\theta = r\, dr\, d\theta — exactly the extra r we measured geometrically before. The mystery factor was the Jacobian all along.

Worked example 2 — a linear transform

Consider the linear map x = 2u + v, y = u + 3v (a shear-and-scale of the plane).

Step 1 — partials are just the coefficients.

x_u = 2, \quad x_v = 1, \quad y_u = 1, \quad y_v = 3.

Step 2 — determinant.

J = \det\!\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} = (2)(3) - (1)(1) = 5.

Step 3 — read off the meaning. A linear map has a constant Jacobian: this transform multiplies every area by |J| = 5. The unit square in uv becomes a parallelogram of area 5 in xy, and any region's area is scaled by the same 5:

\operatorname{area}(R) = 5 \cdot \operatorname{area}(S).

Let T:(u,v) \mapsto (x,y) be a smooth, one-to-one map from a region S onto R, with continuous partial derivatives. Then:

It is no accident that area-scaling and the determinant are the same number — they are defined to be. The determinant of a 2\times2 matrix is, by construction, the signed area of the parallelogram its columns span. The sign records orientation: J > 0 if the map preserves the handedness of the plane, J < 0 if it flips it (a reflection). That is why the integral formula takes the absolute value |J| — area is unsigned, so we discard the orientation bit and keep only the magnitude of the stretch.

This also explains the danger zone: where J = 0, the map collapses a patch to zero area (the two image vectors become parallel — the parallelogram degenerates to a segment). There the change of variables breaks down, which is exactly the origin in polar coordinates, where J = r = 0 and all angles \theta pile up onto a single point.

Watch the area stretch by |J|

The left square lives in the uv-plane (unit area). The right parallelogram is its image under the linear map x = a\,u + v, y = u + a\,v; its area is |J| = a^2 - 1 times the square's. Drag the slider on a and watch the parallelogram — and the printed Jacobian — grow.