Jacobians and general changes of variables

Try

A unit square in parameters (u,v) is mapped by x=2u+v and y=3v. Its corners become (0,0),(2,0),(3,3),(1,3), a parallelogram of area 6. The map stretches and shears the square. A drawing of the four transformed corners explains the shape, while the determinant gives an exact area calculation. A coordinate substitution changes not only the names in the integrand but also how much physical area a parameter patch represents.

Predict

Predict the area scale from the two edge vectors (2,0) and (1,3). Their parallelogram has base 2 and perpendicular height 3, so area 6. The derivative matrix [[2,1],[0,3]] has determinant 6. Swap the order of u and v: the determinant changes sign, but the area remains 6. This is why scalar integration uses an absolute determinant rather than a signed determinant.

Experiment

Integrate f(x,y)=y over the parallelogram. In parameter coordinates f=3v and dA=6 du dv. Over the unit square the integral is ∫₀¹∫₀¹18v du dv=9. Check using the average y-value 3/2 times area 6; the linear density and symmetric parallelogram make that check exact.

Now use the parameter rectangle 0≤u≤2,0≤v≤1. Its image has twice the area and the same y-distribution, so the integral doubles to 18. The determinant is a local scale; the parameter domain supplies how much of that scale is accumulated.

Understand

For a continuously differentiable, one-to-one regular map T on suitable regions, change of variables gives ∫∫R f(x,y)dA=∫∫D f(T(u,v))|det DT(u,v)|du dv, where R=T(D). Standard versions allow boundary exceptions of measure zero. The derivative approximates a small parameter rectangle by a parallelogram, and the determinant measures its area. In three dimensions, the determinant measures local parallelepiped volume.

Three tasks must be checked separately: transform the integrand, transform the region, and include the scaling factor. A correct determinant cannot rescue incorrect bounds. Nonzero determinant gives local invertibility, but global one-to-one coverage still needs examination; a map can fold or wrap a large domain while remaining locally regular on pieces.

Apply

Use x=2u,y=5v to transform x²/4+y²/25≤1 into the unit disk. The determinant is 10, so the ellipse has area 10π. If the integrand is x², it becomes 4u² in addition to the factor 10. The integral is 40 times the unit-disk integral of u², namely 10π. Symmetry gives ∫u²=(1/2)∫(u²+v²)=π/4.

Repair “the Jacobian is negative, so the resulting mass is negative.” A negative determinant records orientation reversal. Scalar mass uses its absolute value. Directed quantities such as flux carry orientation differently; do not transfer the scalar area rule without inspecting the quantity being integrated.

Continue

For optional depth, examine the polar map on a parameter interval twice as long as one revolution. Away from the origin it is locally regular, yet it covers the disk twice. Integrating with its area factor counts both copies. Restrict the angular interval before claiming the integral equals a single-region accumulation.

Create a transformed-region solution that lists boundary equations and checks at least one interior point. This guards against choosing the wrong side of a boundary. Continue to line integrals, where one-dimensional parameter changes use speed as the local length scale, a close relative of the determinant factors developed here.