Change of variables — general Jacobian: Question bank

True or false — justify

The Jacobian determinant is always positive.

False. It carries a sign that records orientation: a positive value means

T

preserves orientation, negative means it flips the plane (like a mirror). We take

∣ J ∣

in integrals precisely because the raw determinant can be negative.

d x d y = r d r d θ

because a polar cell is literally a rectangle of sides

d r

and

r d θ

True. The radial edge has length

d r

, the arc edge has length

r d θ

(arc = radius × angle), so the little wedge is approximately a rectangle of area

r d r d θ

— the same

r

the Jacobian produces.

det J_{T} = 0

at a point, the change-of-variables formula still applies there without issue.

False. A zero Jacobian means the linearized map collapses area to zero (the parallelogram flattens to a segment), so

T

is not locally invertible there. The theorem requires a nonzero Jacobian on the region's interior.

Swapping the roles of

u

and

v

(relabelling) changes the value of the integral.

False. Swapping two columns of the Jacobian matrix flips the sign of

det J

, but the integral uses

∣ J ∣

, so the absolute value — and the answer — is unchanged.

A linear change of variables always has a constant Jacobian over the whole region.

True. For a linear map

x = A u

the partials

\partial x_{i} / \partial u_{j}

are just the entries of the constant matrix

A

, so

det J = det A

is the same everywhere and factors straight out of the integral.

For polar coordinates,

det J = r

can be negative when

r < 0

, so we should write

∣ r ∣

True in principle. If you allow

r < 0

the Jacobian is

r

and area needs

∣ r ∣

; standard convention keeps

r \geq 0

, so

∣ r ∣ = r

and we drop the bars — but only because of that convention.

The change-of-variables formula requires

f

to be positive.

False.

f

can take any sign; it's the area element

∣ J ∣ d u d v

that must stay non-negative, not the integrand. The Jacobian's absolute value handles area;

f

's sign is untouched.

\frac{\partial ( x , y )}{\partial ( u , v )}

and

\frac{\partial ( u , v )}{\partial ( x , y )}

are equal.

False. They are reciprocals: their product is

1

because

J_{T}

and

J_{T^{- 1}}

are inverse matrices and

det J_{T} \cdot det J_{T^{- 1}} = det (I) = 1

Spot the error

"I set

u = x + y

v = x - y

, computed

\frac{\partial ( u , v )}{\partial ( x , y )} = - 2

, and multiplied the integrand by

2

The error is using the wrong-way Jacobian. The integral needs old area per new cell, i.e.

\partial (x, y) / \partial (u, v)

, whose magnitude is

\frac{1}{2}

— the reciprocal of

∣ - 2∣

. Multiplying by

2

overcounts by a factor of

4

"My map flips orientation, so

det J = - 3

, and I multiplied the integrand by

- 3

You forgot the absolute value. Area cannot be negative; you must multiply by

∣ - 3 ∣ = 3

. The minus sign only told you the plane was flipped, which is irrelevant to how much area a cell covers.

"I transformed

f

and the Jacobian correctly but kept the original

x, y

limits of integration."

The region

S

uv

-space was never found. Once you change variables you must re-express the limits in the new coordinates; keeping the old ones integrates over the wrong region entirely.

"For

x = r cos θ, y = r sin θ

I got Jacobian

cos^{2} θ - sin^{2} θ

A sign slipped in the determinant. Correctly,

det (cos θ sin θ - r sin θ r cos θ) = r cos^{2} θ - (- r sin^{2} θ) = r (cos^{2} θ + sin^{2} θ) = r

. The off-diagonal product carries its own minus sign, which cancels.

"The map

T (u, v) = (u^{2}, v)

is fine on

- 1 \leq u \leq 1

because it's smooth."

Smoothness isn't enough —

T

must be injective. Here

u

and

- u

both map to the same

x = u^{2}

, so

T

folds the strip onto itself and double-counts area. You must restrict to

u \geq 0

(or split the region).

"Since the parallelogram edges are

r_{u} d u

and

r_{v} d v

, its area is

∣ r_{u} ∣∣ r_{v} ∣ d u d v

That formula assumes the edges are perpendicular. In general the area is

∣ r_{u} \times r_{v} ∣ d u d v = ∣ x_{u} y_{v} - y_{u} x_{v} ∣ d u d v

, which includes the

sin

of the angle between the edges — exactly the Jacobian.

Why questions

Why is the mapped cell a parallelogram and not some curved blob?

Because we use the first-order (linear) Taylor approximation of

T

near the corner. Under a microscope any smooth map looks linear, and a linear map sends a rectangle to a parallelogram; the curvature is a higher-order effect that vanishes as the cell shrinks.

Why do we use the determinant specifically, rather than some other combination of the partials?

Because the determinant is defined as the signed area/volume-scaling factor of a linear map — see Determinant as Volume Scaling. It's the unique quantity telling you how much a unit cell's area changes under the Jacobian matrix.

Why does the cross product show up in the 2D derivation even though we're in the plane?

Embedding the two edge vectors in 3D as

(a_{x}, a_{y}, 0)

and

(b_{x}, b_{y}, 0)

, their cross product points purely in

z

with magnitude

∣ a_{x} b_{y} - a_{y} b_{x} ∣

— which is the parallelogram's area. It's the cleanest way to extract signed area.

Why can we replace

T

with its linearization at each cell without losing accuracy?

Because we're taking a limit of infinitely fine cells. The error of the linear approximation shrinks faster (order

d u^{2}

) than the cell area (order

d u d v

), so in the integral limit only the linear part survives.

Why does the reciprocal shortcut

\partial (x, y) / \partial (u, v) = 1/ [\partial (u, v) / \partial (x, y)]

hold?

By the multivariable chain rule,

J_{T} J_{T^{- 1}} = I

, so

det J_{T} \cdot det J_{T^{- 1}} = 1

. This is exactly the Inverse Function Theorem guaranteeing the inverse exists where the Jacobian is nonzero.

Why do we choose the change of variables in the first place — what problem is it solving?

To make either the region

R

or the integrand

f

simpler. E.g. Polar Coordinates turn a disk into a rectangle, and

u = x + y

collapses a two-variable integrand into one variable, trading complexity for a Jacobian factor.

Why is

ρ^{2} sin ϕ

the spherical Jacobian and not just

ρ^{2}

The

ρ^{2}

scales the two "sideways" arc-lengths that grow with radius, while

sin ϕ

accounts for the shrinking of longitude circles near the poles — see Spherical and Cylindrical Coordinates. At

ϕ = 0

(the pole)

sin ϕ = 0

and the cell degenerates to a point.

Edge cases

What happens to the polar Jacobian at the origin,

r = 0

It becomes

det J = 0

: every angle

θ

maps to the single point

(0, 0)

, so

T

is not injective there and area collapses. The origin is a measure-zero point, so integrals are unaffected — but

T

is genuinely singular exactly there.

If a region

R

has area zero (a curve or a point), what does the formula give?

Zero. A degenerate region contributes no area, so both sides vanish regardless of

∣ J ∣

. This is why singular points of

T

(isolated zero-Jacobian points) don't spoil the integral.

Can the Jacobian be negative everywhere and the formula still work?

Yes. A globally orientation-reversing map (say a reflection) has

det J < 0

throughout; you just use

∣ J ∣ > 0

and the theorem holds fine. The sign is bookkeeping about orientation, not correctness.

What if

T

is injective on the interior but glues together boundary points (like

θ = 0

and

θ = 2 π

That's fine. The theorem needs injectivity on the interior; boundaries are measure-zero and can overlap without changing the integral. This is why the full disk works despite the seam at

θ = 0

In 3D, what does a zero Jacobian mean geometrically?

The three edge vectors of the mapped cell become coplanar (or collinear), so the parallelepiped flattens to zero volume. The map crushes a 3D neighbourhood into a lower-dimensional set — locally non-invertible.

For the map

u = x^{2} - y^{2}, v = 2 x y

, where does it fail to be invertible?

At the origin

(x, y) = (0, 0)

, where

\partial (u, v) / \partial (x, y) = 4 (x^{2} + y^{2}) = 0

. There the reciprocal

\partial (x, y) / \partial (u, v) = 1/ [4 (x^{2} + y^{2})]

blows up, signalling the Jacobian singularity.

If two different

uv

-cells map to the same

x y

-patch (non-injective

T

), what goes wrong?

You double-count that patch's area, inflating the integral. The formula silently assumes each old cell is covered exactly once; overlaps must be removed by restricting the domain or subtracting.

What is

∣ J ∣

for the identity map

x = u, y = v

, and does that make sense?

∣ J ∣ = 1

: cells don't stretch at all, so

d x d y = d u d v

. It's the sanity-check baseline — no coordinate change, no scaling.

Connections

Change of variables — general Jacobian — the parent this bank drills
Determinant as Volume Scaling — why the determinant is the scaling factor
Cross Product and Area — the area-of-parallelogram engine
Inverse Function Theorem — guarantees the reciprocal shortcut
Polar Coordinates · Spherical and Cylindrical Coordinates — the classic Jacobians