Change of Variables & the Jacobian

From substitution to bending the whole plane

Recall the substitution rule from Volume I — the move you call u-substitution. To integrate a function of x you trade x for a new variable u, and the price of that trade is a factor: dx becomes (dx/du) du. That factor is not decoration. When you re-label the axis, you stretch or compress it, and (dx/du) is exactly how much. Integrate over the right u-interval with that factor in place and the definite integral comes out unchanged, because the area you are actually adding up has been kept honest. Forget the factor and you are measuring the wrong thing.

Now lift that idea into the plane, where this rung lives. In the last two guides you learned to set up a multiple integral as an iterated integral over a region. The trouble is that a region tidy in real life is often hideous in x and y: a disk, an annulus, a slice of pie. Integrating over a disk in Cartesian coordinates means limits that are themselves curved functions, sqrt of this and minus-sqrt of that — correct, but miserable. The cure is to switch to coordinates in which the region becomes a plain rectangle: polar (r, theta) for anything round. But switching from (x, y) to (r, theta) is no innocent re-labelling. It bends the whole plane, and bending distorts area unevenly — a lot near the rim, almost not at all near the center.

So the plane-version of the substitution rule needs a plane-version of the price tag. In one dimension the single number dx/du sufficed because stretching a line is one-dimensional. In two dimensions a tiny patch can be stretched along one direction, squeezed along another, and tilted, all at once — one number cannot capture that. What you need is a single quantity that reports, point by point, how much area a tiny cell gains or loses when it is dragged from the new coordinates into the old. That quantity is the Jacobian determinant, and earning a real picture of it is the whole job of this guide.

The Jacobian: a parallelogram's worth of area

Picture the transformation that sends new coordinates (u, v) to old ones (x, y) = (x(u, v), y(u, v)). Stand at one point and take two tiny steps: one step du in the u-direction, one step dv in the v-direction. In the uv-grid these two steps bound a tiny rectangle of area du dv. But under the map, the u-step lands as some little vector in the xy-plane, and the v-step lands as another little vector — and there is no reason those two image vectors are perpendicular or equal in length. So the tiny uv-rectangle does not map to a tiny xy-rectangle; it maps to a tiny tilted parallelogram spanned by those two image vectors. The map has bent the cell.

What are those two image vectors? Exactly the columns of the Jacobian matrix of the map — the grid of all first partial derivatives, [x_u, x_v; y_u, y_v], where x_u means partial x / partial u and so on. The first column (x_u, y_u) is the velocity of the image point as you push u; the second column (x_v, y_v) is its velocity as you push v. This matrix is just the total derivative of the map: the best linear approximation to the bending, valid in the small neighborhood of the point. The Jacobian matrix tells you the directions; we still need the area.

Here is the one fact that ties it together: the area of the parallelogram spanned by two vectors is the absolute value of the determinant of the matrix with those vectors as columns. So the area of the bent cell is |x_u y_v - x_v y_u| times du dv. That number J = x_u y_v - x_v y_u is the Jacobian determinant — the local area-scaling factor of the map, point by point. If J = 3 near a point, every tiny cell there comes out three times bigger in xy than it was in uv; if J = 1/2, cells shrink by half. In one dimension this determinant collapses to the lone derivative dx/du, so the Jacobian is the genuine, dimension-honest heir of the substitution factor — a determinant standing in for a single derivative.

The theorem, stated honestly

Now the full statement reads like the one-dimensional rule with the determinant in place of the derivative. Suppose a transformation takes new coordinates (u, v) to old ones (x, y), mapping a tidy region S in the uv-plane onto your messy region R in the xy-plane. Then the change-of-variables theorem says: the double integral over R of f(x, y) dA equals the double integral over S of f(x(u, v), y(u, v)) times |J| du dv. In words: rewrite the integrand in the new variables, integrate over the easy region S instead of the hard region R, and insert the absolute Jacobian as the area element. The whole engine of multiple integration runs on this one line, and the same statement holds in three dimensions with a 3-by-3 Jacobian and a volume element |J| du dv dw.

Why does multiplying by |J| at every point give the right total? Because of the same Riemann-sum logic that defines the integral in the first place. Chop S into a fine grid of tiny uv-cells. The map drags each cell over to R as a tiny parallelogram of area |J| du dv. The integral over R is the sum of f times those true xy-areas; written in the new variables, each term is f(x(u,v), y(u,v)) times |J| du dv. Sum and take the limit, and you have the theorem. The Jacobian is not a trick pulled from a hat — it is the honest area of the cell after bending, and the integral is just adding those areas up with the right weights.

Polar coordinates: where the famous r comes from

Let us make it concrete with the change everyone meets first: polar coordinates, x = r cos(theta), y = r sin(theta). You have surely been told that dA = r dr d-theta — that an extra factor of r appears out of nowhere, and that forgetting it silently ruins the answer. The Jacobian explains the r completely, and once you see it you will never forget it. Two ways to see the same thing: compute the determinant, or just look at the geometry of a polar cell.

Map:  x = r cos(theta),  y = r sin(theta)

Jacobian matrix  [ x_r , x_theta ;  y_r , y_theta ]
             =   [ cos(theta) , -r sin(theta) ;
                   sin(theta) ,  r cos(theta) ]

J = x_r y_theta - x_theta y_r
  = cos(theta)(r cos(theta)) - (-r sin(theta))(sin(theta))
  = r cos^2(theta) + r sin^2(theta)
  = r            <-- the famous factor

so   dA = dx dy = |J| dr d-theta = r dr d-theta

The geometry tells the same story without a single derivative, and it is worth holding in your mind's eye. A polar cell is the little patch bounded by two radii at angles theta and theta + d-theta and two arcs at radii r and r + dr. It is not a rectangle of sides dr and d-theta — its inner arc is short and its outer arc is long, because an arc of angle d-theta at radius r has length r d-theta. The cell is a thin curved sliver whose width is dr and whose length is r d-theta, so its area is (r d-theta)(dr) = r dr d-theta. The r is the radius reaching out: cells far from the origin are fat, cells near the origin are thin, and the Jacobian r is precisely that fattening. This is why every polar integral carries an r, and why dropping it is the single most-forgotten symbol in all of multivariable calculus.

This single factor cracks one of the most beautiful integrals in mathematics. The Gaussian integral, the integral from minus infinity to infinity of e^{-x^2} dx, has no elementary antiderivative — but square it, and you get a double integral of e^{-(x^2 + y^2)} over the whole plane. Switch to polar, where x^2 + y^2 = r^2, and the area element hands you exactly the r you need: the integral of e^{-r^2} r dr is elementary because that r is the inside derivative of -r^2/2. Out drops sqrt(pi), so the original integral equals sqrt(pi). The non-elementary one-dimensional integral becomes a one-line two-dimensional computation, and the Jacobian's r is the hinge the whole trick turns on.

A working procedure, and the inverse shortcut

In practice you rarely start with x and y written as functions of u and v. More often you invent a substitution the other way — u and v as combinations of x and y that simplify the region or the integrand — and you do not want to algebraically solve for x and y just to compute J. Here the Jacobian offers a gift: the Jacobian of an inverse map is the reciprocal of the original. So you may compute the determinant of the partials of (u, v) with respect to (x, y), then simply take its reciprocal. The license for this is the inverse function theorem — a non-zero Jacobian at a point certifies the map is locally invertible there — and the reciprocal rule is the multivariable echo of the Volume I fact that dx/du and du/dx are reciprocals. Compute whichever direction is easier.

1. Choose the new coordinates. Pick (u, v) so the region R becomes a rectangle (or simple box) S, or so the integrand collapses. A diamond |x| + |y| <= 1 begs for u = x + y, v = x - y; anything round begs for polar.
2. Find the Jacobian. If x, y are given in terms of u, v, compute J = x_u y_v - x_v y_u directly. If instead u, v are given in terms of x, y, compute that determinant and take its reciprocal.
3. Transform the region. Translate the boundary of R into conditions on u and v; this gives the new limits over S — ideally constants. Watch out for folding or double-covering.
4. Rewrite the integrand and the element. Replace f(x, y) by f expressed in u, v, and replace dA by |J| du dv. Do not forget the absolute value, and do not forget to change the limits to match S.
5. Integrate over S. You now have an ordinary iterated integral over a clean region — evaluate it as you learned in the earlier guides of this rung.

One step up in dimension, the same machine runs. For an orthogonal curvilinear system — cylindrical, spherical, or any in which the coordinate directions meet at right angles — there is a tidy shortcut to the Jacobian: the scale factors. Each scale factor h_i is how much real distance one unit of that coordinate buys, and the volume element is simply their product, dV = h1 h2 h3 du1 du2 du3, which equals |J|. For cylindrical the factors are 1, r, 1, giving dV = r dr d-theta dz; for spherical they are 1, rho, rho sin(phi), giving the celebrated dV = rho^2 sin(phi) d-rho d-phi d-theta. You can always grind out the 3-by-3 determinant, but reading off the scale factors of a familiar system is faster and harder to botch — just remember the product trick only holds when the coordinates are orthogonal.

What to carry forward

Step back and the whole picture is one idea wearing three coats. The Volume I substitution factor dx/du, the polar r, the spherical rho^2 sin(phi) — these are not separate rules to memorize but the same Jacobian determinant, the local stretch factor, computed for different maps. Whenever you change coordinates inside any integral, ask one question: by how much does a tiny cell's area or volume change here? The determinant of the matrix of partial derivatives answers it, and the answer is what you multiply by. Internalize that and the menagerie of memorized area and volume elements collapses into a single understood thing.