Double integrals in polar coordinates — Jacobian r: Question bank

True or false — justify

The area element in polar coordinates is

d r d θ

False. That expression is a length times a dimensionless angle — a length, not an area. The true tile is

r d r d θ

, because the curved side is an arc of length

r d θ

(see Arc Length and Radian Measure).

The factor

r

appears because we chose

r \geq 0

by convention.

False. The

r

is geometric: the arc side of the tile is

r d θ

, so it exists no matter what convention we pick. The

r \geq 0

convention only lets us drop the absolute value

∣ det J ∣ = ∣ r ∣ = r

If a region is a thin square far from the origin, using polar coordinates still forces a factor of

r

into the integral.

True. The

r

comes from the coordinate change, not the region's shape. Even for an off-center square,

d A = r d r d θ

holds — it's just that the limits become messy, which is why we'd avoid polar here.

For the Gaussian integral, dropping the

r

still lets you finish, just with a harder antiderivative.

False. Without

r

you face

\int e^{- r^{2}} d r

, which has no elementary antiderivative. The

r

is exactly what powers the substitution

u = r^{2}

; without it the method collapses (see Gaussian Integral).

The Jacobian for polar coordinates is negative in some quadrants.

False.

det J = r cos^{2} θ + r sin^{2} θ = r \geq 0

everywhere, since

r

is a distance. The trig terms combine to

r

for every

θ

, so no quadrant flips its sign.

Because

θ

is dimensionless, its differential

d θ

carries no units.

True. A radian is arc-length over radius, so it's a pure number. That's precisely why

r d θ

(length

\times

pure number) has units of length and can serve as one side of the area tile.

Increasing

r

by a fixed

d r

always adds the same amount of area to the tile, regardless of where you are.

False. The tile's area is

r d r d θ

, so for the same

d r

and

d θ

the area grows linearly with $r$ — a far tile is fatter than a near one, exactly the pizza-slice picture.

Spot the error

\iint_{R} x d A = \iint_{R^{'}} x r d r d θ

, and

x

is constant so pull it out."

The error is leaving

x

unconverted. You must substitute

x = r cos θ

, giving

\iint (r cos θ) r d r d θ

;

x

is not a constant in

(r, θ)

"Area of the disk

= \int_{0}^{2 π} \int_{0}^{a} d r d θ = 2 π a

Missing the Jacobian. The answer

2 π a

is a length (dimensional alarm), not an area. Restoring

r

gives

\int_{0}^{2 π} \int_{0}^{a} r d r d θ = π a^{2}

"For

x^{2} + y^{2} \leq a^{2}

I'll write

\int_{0}^{a} \int_{0}^{2 π} r d θ d r

with the same limits — order doesn't matter."

Here it happens to work because both limits are constant, but the stated reason is dangerous. Order-swapping is only free when neither limit depends on the other variable; the moment a boundary like

r = 2 cos θ

appears, swapping breaks.

"For the circle

r = 2 cos θ

, integrate

r

from

0

2

and

θ

from

0

2 π

Two errors: the outer

r

-limit is

2 cos θ

(not the constant

2

), and this circle is traced with

θ \in [- \frac{π}{2}, \frac{π}{2}]

, not a full turn — a full sweep double-covers the disk.

d x d y = det J d r d θ

where

det J = r

, so the general rule needs no absolute value."

The general Change of Variables Theorem requires

∣ det J ∣

. It only looks unnecessary here because

r \geq 0

already makes

det J

non-negative; for other maps the modulus is essential.

"Since

det J = r

and

r = 0

at the origin, the substitution is invalid at the origin so the disk integral is wrong."

The Jacobian vanishing on a single point (a set of zero area) does no harm — it contributes nothing to the integral. The origin is a coordinate singularity of measure zero, not a genuine problem.

Why questions

Why does the arc side of the tile use

r d θ

and not just

d θ

Radian measure defines arc length as radius times angle: arc

= r d θ

. The angle alone doesn't know how far out you are; the

r

supplies that distance (see Arc Length and Radian Measure).

Why do we even bother squaring

I = \int e^{- x^{2}} d x

before going polar?

Squaring turns a 1-D integral into a 2-D integral over the whole plane with integrand

e^{- (x^{2} + y^{2})}

. Only then does

x^{2} + y^{2} = r^{2}

appear, letting polar coordinates and the

r

-Jacobian rescue it.

Why does the second-order term

d r d θ \cdot d r

get discarded in the geometric derivation?

The tile's two edges differ by an arc

(r + d r) d θ

r d θ

; the gap is proportional to

d r^{2} d θ

. Compared to the leading

r d r d θ

it shrinks faster as increments

\to 0

, so it vanishes in the limit.

Why does polar tame a disk's limits but not a square's?

A disk's boundary is

r = a

(a constant in polar), giving fixed limits; a square's sides are

x =

const,

y =

const, which become tangled

r

–

θ

relations. Match the coordinate system to the boundary's natural symmetry.

Why must both the integrand and the area element be converted, not just one?

They describe different things: the integrand

f (x, y)

is what you're summing, and

d A

is how much space each sample covers. Converting only one mixes coordinate systems and gives a meaningless quantity.

Why is

r

the same Jacobian whether you derive it by geometry or by determinant?

Both compute the same physical thing — the area-stretch of the coordinate map. The little-box picture and

∣ det J ∣

are two languages for "how much does area scale here," so they must agree (see Jacobian Determinant).

Edge cases

At the origin (

r = 0

), what is the area of a tile, and is that a problem?

The tile area

r d r d θ \to 0

: all angles collapse to one point, so there's genuinely no area there. It's a harmless singularity — a single point contributes nothing to any integral.

θ

ranges over

[0, 4 π]

instead of

[0, 2 π]

for a full disk, what happens?

You sweep the disk twice, doubling the answer to

2 π a^{2}

. Polar coordinates are not unique — you must choose a

θ

-range that covers the region exactly once.

For an annulus

1 \leq r \leq 2

, why are the

θ

-limits still

0

2 π

and constant?

The hole is centered at the origin, so at every angle the same radial slice appears — the boundary has full rotational symmetry. The

r

-limits carry all the "ring" information;

θ

stays a full, constant sweep.

What does the Gaussian integral become if you integrate over only the first quadrant?

Then

θ

runs

0 \to \frac{π}{2}

, giving

I_{quad}^{2} = \frac{π}{2} \cdot \frac{1}{2} = \frac{π}{4}

, i.e. one quarter of

π

— consistent, since one quadrant is a quarter of the plane.

For a circle passing through the origin like

r = 2 cos θ

, what is the tile's area at the single point

θ = π /2

There

r

runs

0 \to 2 cos (π /2) = 0

, so the radial slice has zero length and contributes no area — the curve just grazes the origin at that angle.

If a region dips into

r < 0

under some polar-curve convention, how do we keep

r \geq 0

A negative

r

means "plot in the opposite direction," i.e. add

π

θ

and flip the sign of

r

. Rewriting this way keeps

r \geq 0

so the clean

∣ det J ∣ = r

still holds.

Connections

Double Integrals in Polar Coordinates — Jacobian r — the parent these traps stress-test.
Change of Variables Theorem — the $∣ det J ∣$ that becomes $r$ .
Jacobian Determinant — geometric vs algebraic agreement.
Arc Length and Radian Measure — where $r d θ$ is born.
Gaussian Integral — the killer edge case.
Area of Regions Bounded by Polar Curves — variable $r$ -limits.