core29 min studyLesson 206 in Mathematics

Change of variables and Jacobian determinants

Transform integration regions while accounting for local area or volume scale and multiplicity.

On this page

What you’ll understand

  • Compute a Jacobian determinant as a local volume factor.
  • Transform both the integrand and the integration region.
  • Explain the absolute value and the need to avoid counting the image repeatedly.

A circular region is awkward in rectangular coordinates but simple in radius and angle. The convenience comes with a price: equal coordinate rectangles need not represent equal physical areas. A change of variables must account for that stretching, together with any overlap in how the coordinates cover the region. The Jacobian determinant supplies the local scale. Understanding where this factor comes from makes coordinate substitution a geometric calculation rather than a memorized list of special formulas.

A derivative turns small boxes into parallelotopes

Near a point u, a differentiable map T behaves to first order like its derivative DT(u). A small coordinate box therefore becomes approximately a parallelotope whose edge vectors are the derivative's columns multiplied by the corresponding coordinate widths. The absolute determinant of that derivative is its volume scale. The determinant's sign records whether orientation is preserved or reversed; ordinary unsigned volume uses the magnitude. A negative determinant does not mean negative area, although it matters in oriented integration of differential forms.

For a standard sufficient version of the substitution theorem, take a C¹ diffeomorphism between open subsets of equal-dimensional Euclidean space, and integrate a nonnegative measurable function or an absolutely integrable function. The transformed integrand is the original function composed with T, multiplied by the absolute Jacobian determinant. Restricting to a measurable subregion gives its transformed version. Many familiar coordinate systems have exceptional boundaries or axes where they are not diffeomorphisms; those sets can be excluded when they have measure zero and the remaining hypotheses hold.

∫T(U)f(x) dx=∫Uf(T(u)) ∣det⁡DT(u)∣ du\int_{T(U)} f(x)\,dx=\int_U f(T(u))\,|\det DT(u)|\,du
Here T is a C¹ diffeomorphism on the open coordinate region U; f is nonnegative measurable or absolutely integrable. Appropriate measurable subregions can be included with indicator functions.↔ Scroll wide equations horizontally.

Polar coordinates illustrate both scale and coverage

In the plane, T(r,θ)=(r cos θ,r sin θ). Its derivative columns point in the radial and angular directions. Their lengths are one and r, and they are perpendicular, giving determinant r. Thus a small angular interval has physical length approximately r times its angle width. Using an angular interval of length one full turn covers an annulus once except for a seam; using two full turns covers almost every point twice. The factor r corrects local scale but does not correct this global double coverage by itself.

For a disk, the origin and the chosen angular seam cause failures of the simplest diffeomorphism description. They have area zero, so one may work on the punctured disk with a seam removed and then restore those null sets. This explanation is more precise than claiming polar coordinates are globally one-to-one on the entire closed rectangle of radius and angle. If an integrand is singular at the origin, nullity of the point does not guarantee finite mass nearby; its transformed radial behavior must still be checked for integrability.

DT(r,θ)=(cos⁡θ−rsin⁡θsin⁡θrcos⁡θ),det⁡DT=r,dA=r dr dθDT(r,\theta)=\begin{pmatrix}\cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta\end{pmatrix},\qquad \det DT=r,\qquad dA=r\,dr\,d\theta
The factor r is positive for r>0. The standard full-disk formula follows by handling the origin and one seam as null exceptional sets.↔ Scroll wide equations horizontally.

A useful substitution changes the geometry as well as the formula

A substitution is valuable when it simplifies a region, an integrand, or both. For an ellipse, scaling a unit disk by its semiaxes converts a curved inequality into a round one and contributes a constant area factor. For a linear shear, the image can be visually slanted while its area remains unchanged when the determinant is one. This is why judging scale from the longest visible edge is unreliable: stretching one direction, shrinking another, and shear combine through the full determinant.

A reliable calculation writes the coordinate map, its valid region, the transformed bounds, and its determinant before integrating. Check dimensions and multiplicity separately. A map with a vanishing determinant cannot satisfy the ordinary diffeomorphism theorem at that point, but may still be usable after partitioning into regular regions or applying a more general result. The theorem is not permission to discard singular sets without analysis. In higher dimensions the same principle gives cylindrical and spherical volume factors, derived from their own Jacobian matrices.

Mass over an ellipse

Evaluate the integral of x²+y² over the filled ellipse x²/4+y²≤1.

  1. Use x=2r cosθ and y=r sinθ, with 0≤r≤1 and one full angular turn. This is a scaling of polar coordinates, with absolute Jacobian 2r away from the usual null exceptions.

  2. The integrand becomes r²(4cos²θ+sin²θ). Including area gives 2r³(4cos²θ+sin²θ), which separates into radial and angular factors.

  3. The radial integral of 2r³ from 0 to 1 is 1/2. Each of the full-turn integrals of cos²θ and sin²θ is π, so the angular factor is 5π.

The result

The integral is 5π/2. Omitting the Jacobian would change both the radial weighting and the physical meaning of the accumulation.

Explore the model

Interactive lab

Interactive lab · matrix

Shear preserves area while scale changes it

Reset, set a=1 and b=1.5, and predict area scale 1 even though the square slants. For an independent run, Reset, set a=0.5 and b=1.5: predict area scale 0.5. The instrument shows the triangular linear map [[a,b],[0,1]]; it illustrates a constant Jacobian, not a spatially varying nonlinear coordinate map.

Shear preserves area while scale changes itThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe determinant measures area scale even when shear changes shape without changing that area.

Check your understanding

Why does an ordinary scalar area integral use |det DT| rather than det DT?

Try it yourself

Find the integral of 1/(x²+y²) over the annulus 1≤√(x²+y²)≤2. Explain why the same integrand on the punctured unit disk has infinite integral.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

After including the polar area factor, identify the remaining radial power.

Compare with the solution

On the annulus the integrand is r⁻² and area is r dr dθ, giving 2π∫₁² dr/r=2π log 2. On 0<r≤1, the corresponding integral is 2π∫₀¹ dr/r, which diverges to positive infinity. Removing the single origin does not control the accumulating mass in every neighborhood of it. Tonelli allows the latter nonnegative integral to be interpreted as infinite.

Take it with you

  • A Jacobian determinant measures local volume scale; its sign separately records orientation.
  • Transform the integrand, region, and area factor together.
  • Null exceptional sets and multiple coverage require explicit justification rather than visual assumptions.

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