📐 Concept diagram

07-03 — Change of Variables and Jacobians

Phase: 7 — Calculus IV: Vector Calculus Subject: 07-03 Prerequisites: 07-02 — Triple Integrals in Cylindrical and Spherical Coordinates, double integrals (Phase 6) Next subject: 07-04 — Vector Fields

Learning Objectives

By the end of this subject, you will be able to:

1. State the general change-of-variables formula for double and triple integrals and explain the role of the Jacobian determinant
2. Compute Jacobian determinants for 2D and 3D transformations, including polar, cylindrical, spherical, and general affine transformations
3. Use the Jacobian to determine how area/volume elements scale under a coordinate transformation
4. Apply the change-of-variables formula to simplify integrals over nonstandard regions by mapping them to rectangles or boxes
5. Recognize and handle transformations where the Jacobian is zero (degenerate maps) and understand the requirement that the transformation be one-to-one on the interior

Core Content

1. The Change-of-Variables Formula (General)

⚠️ CRITICAL FOUNDATION: The Jacobian determinant |∂(x,y,z)/∂(u,v,w)| is the volume-scaling factor for coordinate transformations. dA = |J| du dv in 2D, dV = |J| du dv dw in 3D. Polar, cylindrical, and spherical coordinates are special cases.

For a transformation T from the uv-plane to the xy-plane:

$T(u, v) = (x(u, v), y(u, v))
$

The change-of-variables formula for double integrals is:

$∬_R f(x, y) dx dy = ∬_S f(x(u, v), y(u, v)) · |J(u, v)| du dv
$

Where S is the region in the uv-plane that maps to R in the xy-plane, and J is the Jacobian determinant:

$        |∂x/∂u  ∂x/∂v|
J(u,v) =|∂y/∂u  ∂y/∂v| = (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)
$

For triple integrals (transformation from uvw-space to xyz-space):

$∭_E f(x, y, z) dx dy dz = ∭_T f(x(u,v,w), y(...), z(...)) · |J(u,v,w)| du dv dw
$

Where:

$        |∂x/∂u  ∂x/∂v  ∂x/∂w|
J = det |∂y/∂u  ∂y/∂v  ∂y/∂w|
        |∂z/∂u  ∂z/∂v  ∂z/∂w|
$

Key requirement: The transformation T must be one-to-one (injective) on the interior of S (it can be many-to-one on the boundary, a set of measure zero). The Jacobian must not change sign in the interior (or we use |J|).

Intuition: The factor |J| du dv is the area of the small "parallelogram" in the xy-plane that corresponds to the small rectangle du × dv in the uv-plane. It measures how much the transformation stretches or compresses area.

2. The Jacobian as Area/Volume Scaling Factor

Consider a small rectangle in the uv-plane of size Δu × Δv at (u₀, v₀). Under the transformation T, this maps approximately to a parallelogram in the xy-plane spanned by the vectors:

$v₁ = T(u₀+Δu, v₀) − T(u₀, v₀) ≈ (∂x/∂u, ∂y/∂u) Δu
v₂ = T(u₀, v₀+Δv) − T(u₀, v₀) ≈ (∂x/∂v, ∂y/∂v) Δv
$

The area of this parallelogram is |v₁ × v₂| = |det([v₁ v₂])| = |J| Δu Δv.

So the local area scaling factor is exactly |J|.

Example — Polar coordinates: x = r cos θ, y = r sin θ.

$    |∂x/∂r  ∂x/∂θ|   |cos θ  −r sin θ|
J = |∂y/∂r  ∂y/∂θ| = |sin θ   r cos θ| = r cos²θ + r sin²θ = r
$

So dA = r dr dθ. The area scaling grows linearly with r — a small angular sector at larger r covers more area.

Example — Scaling transformation (linear): x = au, y = bv.

$    |a  0|
J = |0  b| = ab
$

Area scales by |ab|: a rectangle of area ΔuΔv maps to a rectangle of area |ab|ΔuΔv.

3. The Jacobian for Common 3D Transformations

Cylindrical: x = r cos θ, y = r sin θ, z = z.

$    |cos θ  −r sin θ  0|
J = |sin θ   r cos θ  0| = r(cos²θ + sin²θ) = r
    |  0       0      1|
$

Spherical: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.

$    |sin φ cos θ  ρ cos φ cos θ  −ρ sin φ sin θ|
J = |sin φ sin θ  ρ cos φ sin θ   ρ sin φ cos θ| = ρ² sin φ
    |cos φ       −ρ sin φ          0          |
$

General linear (affine) transformation: x = a₁u + b₁v + c₁w, etc.

$J = det(A) where A is the transformation matrix.
$

For a pure scaling diag(a, b, c): J = abc. For a rotation: J = 1 (rotations preserve volume). For a shear: J = 1 (shear preserves volume).

4. Using Change of Variables to Simplify Integrals

The strategy is: 1. Identify a transformation (u, v) = g(x, y) that maps the region to a simple shape 2. Find the inverse transformation (x, y) = T(u, v) 3. Compute the Jacobian J = ∂(x,y)/∂(u,v) 4. Express the integrand and bounds in u, v 5. Evaluate the new integral (usually much simpler)

Example 1: Evaluate ∬_R (x + y) dA where R is the parallelogram with vertices (0,0), (2,1), (1,3), (−1,2).

This is a parallelogram. We can map it to the unit square using a linear transformation.

Let's find the transformation. The parallelogram is spanned by vectors:
v₁ = (2,1) and v₂ = (1,3).

Any point in R is (x,y) = u(2,1) + v(1,3) for u, v ∈ [0, 1].
So the transformation is T(u, v) = (2u + v, u + 3v).

Jacobian: J = |2  1| = 6 − 1 = 5.
              |1  3|

Integrand: x + y = (2u+v) + (u+3v) = 3u + 4v.

∬_R (x+y) dA = ∫₀¹ ∫₀¹ (3u + 4v) · |5| du dv
= 5 ∫₀¹ [3u²/2 + 4vu]₀¹ dv
= 5 ∫₀¹ (3/2 + 4v) dv
= 5 [3v/2 + 2v²]₀¹ = 5(3/2 + 2) = 5(7/2) = 35/2.

Example 2: Evaluate ∬_R e^{(y−x)/(y+x)} dA where R is the triangle with vertices (0,0), (1,0), (0,1).

The integrand and region suggest the substitution:
u = y − x,  v = y + x

Then:
x = (v − u)/2,  y = (v + u)/2

Jacobian:
∂x/∂u = −1/2,  ∂x/∂v = 1/2
∂y/∂u = 1/2,   ∂y/∂v = 1/2

J = (−1/2)(1/2) − (1/2)(1/2) = −1/4 − 1/4 = −1/2
|J| = 1/2.

Now find the region in uv-coordinates.
Triangle vertices:
(0,0) → u=0, v=0
(1,0) → u=−1, v=1
(0,1) → u=1, v=1

The triangle maps to: u from −v to v, v from 0 to 1.

Actually, let's sketch: (0,0), (−1,1), (1,1). This is a triangle with base on v=1 from u=−1 to u=1, and apex at (0,0). So:
v ∈ [0, 1], u ∈ [−v, v].

∬ e^{u/v} · (1/2) du dv
= (1/2)∫₀¹ ∫_{−v}^v e^{u/v} du dv

Inner: ∫_{−v}^v e^{u/v} du. Let w = u/v, du = v dw.
= v ∫₋₁¹ e^w dw = v(e − e^{−1}) = v(e − 1/e)

= (1/2)(e − 1/e) ∫₀¹ v dv = (1/2)(e − 1/e)(1/2) = (e − 1/e)/4.

5. Transformations for Nonstandard Regions

Elliptical regions: x²/a² + y²/b² ≤ 1.

Use: x = a r cos θ, y = b r sin θ (scaled polar). Jacobian: J = a b r. Region: r ∈ [0, 1], θ ∈ [0, 2π].

Then: ∬_R f(x, y) dA = ∫₀^{2π} ∫₀¹ f(a r cos θ, b r sin θ) · a b r dr dθ.

Example 3: Find the area of the ellipse x²/4 + y²/9 ≤ 1.

$x = 2r cos θ, y = 3r sin θ. J = 6r.
Area = ∬ 1 dA = ∫₀^{2π} ∫₀¹ 1 · 6r dr dθ = 2π · 6 · [r²/2]₀¹ = 6π.
(Check: πab = π·2·3 = 6π. ✓)
$

6. When the Jacobian is Zero

If J(u, v) = 0 at some point, the transformation is locally degenerate — it collapses area/volume to zero at that point. This is acceptable if it happens on a set of measure zero (like at the origin in polar coordinates, where r = 0 gives J = 0).

However, if J = 0 over a region of positive measure, the transformation is not one-to-one, and the formula doesn't apply directly (you'd need to partition the domain).

Example J=0: T(u, v) = (u², v). J = 2u. At u = 0, the transformation is not invertible (both u and −u map to u²). This transformation is not one-to-one on any region symmetric about u=0, and the change-of-variables formula would need special handling.

Common Misconceptions

1. "The Jacobian is always positive." The Jacobian can be negative; the formula uses |J|. A negative Jacobian means the transformation reverses orientation.

2. "Any substitution works like 1D u-substitution." In multiple dimensions, you can't just substitute expressions — you need a valid coordinate transformation that maps the region and you must include the Jacobian.

3. "dx dy = du dv always." No — the volume element changes. Only with |J| = 1 (rotations, translations, shears) is dx dy = du dv.

4. "The Jacobian is the derivative of the transformation." It's the determinant of the derivative matrix (the matrix of partial derivatives). In 1D, the change-of-variables formula reduces to ∫ f(g(u))|g'(u)| du, where |g'| is the 1D Jacobian.

Key Terms

• For triple integrals
• Jacobian determinant

Worked Examples

Example 1: A Nonlinear Transformation

Problem: Evaluate ∬_R xy dA where R is the region in the first quadrant bounded by y = x, y = 3x, xy = 1, and xy = 4.

Solution:

The boundaries suggest: u = y/x, v = xy.

Boundaries in uv:
y = x   → u = 1
y = 3x  → u = 3
xy = 1  → v = 1
xy = 4  → v = 4

Region S: u ∈ [1, 3], v ∈ [1, 4] — a rectangle!

Find the inverse transformation (solve for x, y):
v = xy, u = y/x → ux² = yx · (y/x implicitly...)
Actually: y = ux, and v = x(ux) = ux², so x² = v/u → x = √(v/u), y = u√(v/u) = √(uv).

Jacobian: ∂(x,y)/∂(u,v).
Compute: x = v^{1/2} u^{−1/2}, y = u^{1/2} v^{1/2}.

∂x/∂u = (−1/2) v^{1/2} u^{−3/2}
∂x/∂v = (1/2) v^{−1/2} u^{−1/2}
∂y/∂u = (1/2) u^{−1/2} v^{1/2}
∂y/∂v = (1/2) u^{1/2} v^{−1/2}

J = (−1/2 v^{1/2} u^{−3/2})(1/2 u^{1/2} v^{−1/2}) − (1/2 v^{−1/2} u^{−1/2})(1/2 u^{−1/2} v^{1/2})
  = −1/4 u^{−1} − 1/4 u^{−1} = −1/(2u).

|J| = 1/(2u).

Integrand: xy = v.

∬_R xy dA = ∫₁³ ∫₁⁴ v · (1/(2u)) dv du
= (1/2) ∫₁³ (1/u) du ∫₁⁴ v dv
= (1/2) [ln u]₁³ · [v²/2]₁⁴
= (1/2)(ln 3)((16−1)/2)
= (1/2)(ln 3)(15/2)
= (15/4) ln 3.

Example 2: Double Integral with Complicated Integrand

Problem: Evaluate ∬_R sin(x + 2y) cos(x − 2y) dA over the square [0, π/2] × [0, π/2].

Solution:

Use: u = x + 2y, v = x − 2y.

Solve for x, y:
x = (u + v)/2, y = (u − v)/4.

Jacobian:
∂x/∂u = 1/2, ∂x/∂v = 1/2
∂y/∂u = 1/4, ∂y/∂v = −1/4

J = (1/2)(−1/4) − (1/2)(1/4) = −1/8 − 1/8 = −1/4.
|J| = 1/4.

Find the new region S. Original: x ∈ [0, π/2], y ∈ [0, π/2].
In uv:
u = x + 2y ∈ [0, π/2 + π] = [0, 3π/2]
v = x − 2y ∈ [−π, π/2]

But these aren't independent — we need the image of the rectangle. The map is linear; the rectangle maps to a parallelogram. Bounds:
u+v = 2x ∈ [0, π] → u+v ∈ [0, π]
u−v = 4y ∈ [0, 2π] → u−v ∈ [0, 2π]

So region S is the parallelogram:
max(0, v) ≤ u ≤ π − v (from u+v ≤ π and u ≥ 0, u ≥ −v? let's work systematically).

Actually, with a linear map from a rectangle, the image is a parallelogram. The vertices:
(0,0) → (0,0)
(π/2,0) → (π/2, π/2)
(π/2,π/2) → (3π/2, −π/2)
(0,π/2) → (π, −π)

This is messy to describe. Alternative: the integrand sin(u)cos(v) factors, so let's integrate in the original coordinates directly using trig identity:

sin(x+2y)cos(x−2y) = (1/2)[sin((x+2y)+(x−2y)) + sin((x+2y)−(x−2y))]
= (1/2)[sin(2x) + sin(4y)]

Now: ∫₀^{π/2} ∫₀^{π/2} (1/2)[sin(2x) + sin(4y)] dx dy
= (1/2)∫₀^{π/2} ∫₀^{π/2} sin(2x) dx dy + (1/2)∫₀^{π/2} ∫₀^{π/2} sin(4y) dx dy
= (1/2)(π/2)[−cos(2x)/2]₀^{π/2} + (1/2)(π/2)[−cos(4y)/4]₀^{π/2}
= (π/4)[(−(−1)+1)/2] + (π/4)[(−1+1)/4]
= (π/4)(1) + 0 = π/4.

This demonstrates: sometimes a trigonometric identity is better than a change of variables!

Example 3: Volume Under a Paraboloid via Affine Transformation

Problem: Find the volume under z = x² + y² above the elliptical region x²/4 + y²/9 ≤ 1.

Solution:

$Use scaled polar: x = 2r cos θ, y = 3r sin θ. J = 6r.
Region: r ∈ [0, 1], θ ∈ [0, 2π].

Integrand: x² + y² = 4r² cos²θ + 9r² sin²θ = r²(4 cos²θ + 9 sin²θ).

V = ∫₀^{2π} ∫₀¹ r²(4 cos²θ + 9 sin²θ) · 6r dr dθ
= 6 ∫₀^{2π} (4 cos²θ + 9 sin²θ) dθ · ∫₀¹ r³ dr
= 6 · (1/4) · ∫₀^{2π} (4 cos²θ + 9 sin²θ) dθ

∫₀^{2π} cos²θ dθ = π,  ∫₀^{2π} sin²θ dθ = π.

= (6/4)(4π + 9π) = (3/2)(13π) = 39π/2.
$

Quiz

Q1: The change-of-variables formula for a transformation T(u, v) → (x, y) replaces dA with:

A) du dv B) |∂(x, y)/∂(u, v)| du dv C) (∂x/∂u + ∂y/∂v) du dv D) √(∂x/∂u · ∂y/∂v) du dv

Correct: B)

• If you chose B: dA = |J| du dv where J = det[∂x/∂u ∂x/∂v; ∂y/∂u ∂y/∂v] is the Jacobian determinant. Correct!
• If you chose A: Forgetting the Jacobian — only true if the transformation preserves area (J = ±1).
• If you chose C: That's the trace of the Jacobian matrix, not the determinant.
• If you chose D: The Jacobian is a determinant, not a product of square roots.

Q2: For polar coordinates x = r cos θ, y = r sin θ, the Jacobian ∂(x, y)/∂(r, θ) equals:

A) 1 B) r C) cos θ + sin θ D) r²

Correct: B)

• If you chose B: det[[cos θ, −r sin θ], [sin θ, r cos θ]] = r cos²θ + r sin²θ = r. Correct!
• If you chose A: Forgetting the r that comes from differentiating r cos θ and r sin θ.
• If you chose C: This is not the determinant of the Jacobian matrix.
• If you chose D: The Jacobian is r, not r².

Q3: The Jacobian is zero at a point. What does this indicate?

A) The integral can't be computed B) The transformation is not one-to-one near that point (fails to be locally invertible) C) The region has zero area D) The integral is zero

Correct: B)

• If you chose B: A zero Jacobian means the transformation collapses dimensions locally — it's not invertible. Change of variables requires an injective map on the interior. Correct!
• If you chose A: The integral can still be computed if the set where J = 0 has measure zero.
• If you chose C: The Jacobian zero means local area scaling is zero, not that the original region has zero area.
• If you chose D: The integral depends on the integrand, not just the Jacobian.

Q4: The change-of-variables formula for triple integrals replaces dV with:

A) |J| du dv dw B) J du dv dw (without absolute value) C) |J|² du dv dw D) (1/|J|) du dv dw

Correct: A)

• If you chose A: dV = |∂(x,y,z)/∂(u,v,w)| du dv dw — absolute value of the 3×3 Jacobian determinant. Correct!
• If you chose B: The absolute value is essential — volume elements are positive regardless of orientation.
• If you chose C: The volume element scales by |J|, not |J|².
• If you chose D: That's the inverse — dV scales by |J|, not 1/|J|.

Q5: Which transformation maps the unit square [0,1]×[0,1] to the parallelogram with vertices (0,0), (2,1), (3,3), (1,2)?

A) x = u + v, y = u + 2v B) x = 2u + v, y = u + 2v C) x = u, y = v D) x = 2u, y = v

Correct: B)

• If you chose B: Check: (u,v)=(0,0)→(0,0); (1,0)→(2,1); (0,1)→(1,2); (1,1)→(3,3). |J| = 2·2−1·1 = 3. Correct!
• If you chose A: (1,0)→(1,1) not (2,1) — wrong.
• If you chose C: This is the identity — maps to a unit square, not the parallelogram.
• If you chose D: Only stretches x, doesn't create the slanted shape.

Q6: For spherical coordinates (x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ), the Jacobian |∂(x,y,z)/∂(ρ,φ,θ)| equals:

A) ρ B) ρ² C) ρ² sin φ D) ρ² cos φ

Correct: C)

• If you chose C: The Jacobian determinant evaluates to ρ² sin φ. This is why dV = ρ² sin φ dρ dφ dθ. Correct!
• If you chose A: That's the cylindrical/polar Jacobian, not spherical.
• If you chose B: Missing the sin φ factor that accounts for latitude.
• If you chose D: The factor is sin φ, not cos φ.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Compute the Jacobian for the transformation x = u² − v², y = 2uv.

Problem 2: Evaluate ∬_R (x − y) dA where R is the parallelogram with vertices (0,0), (3,0), (4,2), (1,2).

Problem 3: Use the substitution u = x + y, v = x − y to evaluate ∬_R (x² − y²) dA where R is the square [0, 1] × [0, 1].

Problem 4: Find the area of the region bounded by the curves xy = 1, xy = 4, y = x, y = 2x in the first quadrant. (Use u = xy, v = y/x.)

Problem 5: Evaluate ∬_R e^{x²+y²} dA where R is the unit disk x² + y² ≤ 1.

Problem 6: Find the volume inside the ellipsoid x²/a² + y²/b² + z²/c² = 1 using the transformation x = au, y = bv, z = cw.

Problem 7: Evaluate ∬_R (x + y)² sin²(x − y) dA over the square with vertices (π,0), (2π,π), (π,2π), (0,π). (Hint: u = x + y, v = x − y.)

Summary

1. The change-of-variables formula ∬_R f dx dy = ∬_S f(T(u,v)) · |J| du dv generalizes 1D u-substitution to multiple dimensions; the Jacobian determinant |J| measures local area/volume scaling
2. The Jacobian is the determinant of the derivative matrix of the transformation; for polar it's r, for cylindrical it's r, for spherical it's ρ² sin φ
3. A well-chosen transformation maps a complicated region to a simple rectangle/box and simultaneously simplifies the integrand — the art is finding the right substitution
4. For affine transformations (scaling, rotation, shear), the Jacobian is constant; for nonlinear transformations, it varies and must be included inside the integral
5. The transformation must be one-to-one on the interior of the domain; J = 0 on a set of measure zero (e.g., r = 0 in polar) is acceptable, but J = 0 on a region requires partitioning

Pitfalls

• Forgetting the absolute value of the Jacobian. The change-of-variables formula requires |J|, not J. If J is negative, using J instead of |J| can produce a negative volume/area for a positive integrand — a clear sign of error. The absolute value ensures the area/volume element is always positive.
• Assuming dx dy = du dv. In multiple dimensions, the differential area/volume element CHANGES under coordinate transformation. Only when |J| = 1 (rotations, translations, shears with determinant ±1) can you simply replace dx dy with du dv. Every other transformation requires multiplying by |J|.
• Using a non-one-to-one transformation without partitioning. The change-of-variables theorem requires the transformation to be injective on the interior of the domain. Transformations like x = u², y = v are not one-to-one on domains symmetric about u = 0. Either restrict the domain to where the transformation is injective, or partition and handle each piece separately.
• Computing the wrong Jacobian. The Jacobian needed is ∂(x,y)/∂(u,v) — the determinant of partial derivatives of the ORIGINAL variables with respect to the NEW variables. Computing ∂(u,v)/∂(x,y) by mistake requires inverting the result (since ∂(x,y)/∂(u,v) = 1 / ∂(u,v)/∂(x,y)). Using the wrong one produces the reciprocal factor.
• Treating the change of variables like 1D u-substitution. In 1D, ∫ f(g(u)) g'(u) du works directly. In 2D/3D, you cannot simply substitute expressions and multiply by some derivative — you need a complete coordinate transformation that maps the region, and you must compute the full Jacobian determinant. Piecewise or ad hoc substitutions that work in 1D can silently produce wrong results in higher dimensions.

Next Steps

Move on to 07-04 — Vector Fields to learn about vector fields in 2D and 3D, gradient fields, potential functions, and the fundamental concepts that connect integration to vector calculus.