Triple integrals in Cartesian, cylindrical, spherical coordinates: Question bank

First, three words we lean on everywhere below, each anchored to a picture in your head:

Here $r$ is distance from the $z$ -axis, $ρ$ is distance from the origin, $θ$ is the angle around the $z$ -axis, and $ϕ$ is the angle down from the $+ z$ -axis. Keep those four pictures in mind — every trap below is really about one of these boxes.

True or false — justify

True or false: If

f = 1

, then

∭_{E} f d V

gives the surface area of

E

False — it gives the volume of

E

. Surface area is a 2D integral over the boundary of

E

, not a 3D integral over its interior.

True or false: In cylindrical coordinates you can drop the

r

factor whenever the integrand has no

r

in it.

False — the

r

comes from the volume element (the fan-shaped wedge widening with

r

), not from the integrand. It stays regardless of what

f

looks like.

True or false: The Jacobian factor is always positive.

True in the formula, because we take the absolute value

∣ J ∣

. The raw determinant can be negative (orientation-reversing), but volume is never negative, so we strip the sign.

True or false: For a ball, spherical coordinates turn the region into a rectangular box in

(ρ, ϕ, θ)

-space.

True — the ball

ρ \leq a

becomes

0 \leq ρ \leq a, 0 \leq ϕ \leq π, 0 \leq θ < 2 π

, all constant limits. That constant-limit "box" is exactly why spherical is so powerful for round regions.

True or false: In spherical coordinates the angle

ϕ

can range over

[0, 2 π)

just like

θ

False —

ϕ \in [0, π]

only. It measures tilt from the north pole to the south pole; going past

π

would re-cover points you already reached with a different

θ

True or false: The order

d z d y d x

and

d x d y d z

always give the same number for a fixed region.

True — the value of a convergent integral is order-independent (Fubini). But the limits usually change completely, and only some orders are easy to set up.

True or false:

d V = ρ^{2} sin ϕ d ρ d ϕ d θ

can be replaced by

ρ^{2} sin θ d ρ d ϕ d θ

when the region is symmetric.

False — it is never

sin θ

. The shrinking that the sine encodes happens at the poles (

ϕ = 0, π

), which is governed by

ϕ

, not by the around-axis angle

θ

True or false: Cylindrical coordinates are just polar coordinates with a

z

stacked on top.

True — that's exactly the picture. See Double integrals & polar coordinates:

r d r d θ

is the polar area element, and multiplying by

d z

lifts it into 3D.

Spot the error

A student writes

∭_{E} d V = \int_{0}^{1} \int_{0}^{x} \int_{0}^{x y} d z d y d x

and says the outer limits are

0

x

. Error?

The outer

x

-limits must be constants, but here they'd depend on

x

itself — nonsense. By the time you reach the outer integral, all inner variables are gone, so only a number can remain.

A student uses

d V = r d r d θ d z

but lets

r

range over

[- a, a]

to cover the whole disk. Error?

In cylindrical coordinates

r \geq 0

always; negative

r

would double-count points. The full disk is covered by

r \in [0, a]

together with

θ \in [0, 2 π)

A student converts

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

to spherical and writes the boundary as

ρ^{2} = a^{2}

, then integrates

ρ

from

- a

a

. Error?

ρ

is a distance, so

ρ \geq 0

; it runs

0

a

, not

- a

a

. The other side of the ball is reached by the angles, not by negative

ρ

A student computes the cylinder volume as

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

(no

r

) and gets

2 π ah

. Error?

They dropped the Jacobian

r

. The correct integrand is

r

, giving

2 π \cdot \frac{a ^{2}}{2} \cdot h = π a^{2} h

. Missing

r

produces the wrong units and wrong number.

A student writes the Jacobian as a sum of the partial derivatives instead of the determinant of the matrix. Error?

The local volume stretch is the determinant of the derivative matrix (signed volume of the parallelepiped its columns span), not a sum. See Jacobian and change of variables.

A student integrates

\int_{0}^{π} sin ϕ d ϕ

and gets

0

"because sine is symmetric". Error?

Over

[0, π]

sin ϕ \geq 0

throughout, so the integral is

2

, not

0

. The cancellation intuition applies to

sin

over

[0, 2 π]

, a different range that

ϕ

never uses.

A student projects a hemisphere onto the

x y

-plane and integrates

z

from

0

a

(a constant). Error?

The top surface is the curved sphere

z = a^{2} - x^{2} - y^{2}

, which depends on

x, y

— not the constant

a

. Only the flat base is at a constant height.

Why questions

Why does the cylindrical volume element carry an extra

r

but Cartesian carries nothing?

A Cartesian brick is the same size everywhere, so no correction is needed. A cylindrical wedge fans out, so its arc-length

r d θ

grows with

r

— the

r

records that growth.

Why is there a

ρ^{2}

(two powers) in spherical but only one

r

in cylindrical?

Spherical has two curved arc-edges (

ρ d ϕ

and

ρ sin ϕ d θ

), each contributing one

ρ

. Cylindrical has only one curved edge (

r d θ

), so only one factor.

Why does

sin ϕ \to 0

at the poles make physical sense?

ϕ = 0

ϕ = π

the latitude circles collapse to a single point, so the

θ

-arc has zero length. The

sin ϕ

factor shrinks the box to zero exactly there, preventing over-counting.

Why must you integrate inside-out (innermost limits can depend on outer variables, never the reverse)?

Each integration "uses up" its variable; once integrated, that variable is gone. So an inner limit may still see the outer variables (not yet integrated), but an outer limit cannot see an inner variable that no longer exists.

Why prefer spherical the moment you see

x^{2} + y^{2} + z^{2}

in a region or integrand?

Because

x^{2} + y^{2} + z^{2} = ρ^{2}

collapses to a single clean variable, turning ugly nested square-root limits into a constant boundary

ρ = a

Why is a determinant — and not, say, an average of the edge lengths — the right measure of local stretch?

A determinant is the signed volume of the parallelepiped spanned by the column (edge) vectors, which is precisely the warped-box volume we need. Averaging lengths would ignore the angles between edges.

Why does the ball-volume integral factor into three separate 1D integrals?

All limits are constants and the integrand

ρ^{2} sin ϕ

splits as (function of

ρ

)×(function of

ϕ

)×

1

. Fubini then lets a product of separable factors become a product of independent integrals.

Edge cases

What is

∭_{E} f d V

when

E

has zero volume (e.g. a flat disk sitting in 3D)?

It is

0

— a 2D region has no 3D volume, so every

Δ V_{k}

collapses and the sum vanishes, whatever

f

is.

At the origin

ρ = 0

, spherical coordinates are degenerate — why is that harmless for the integral?

ρ = 0

the volume element

ρ^{2} sin ϕ

is zero, so that single degenerate point (and the whole polar axis) contributes nothing. The ambiguity in

θ, ϕ

there never affects the answer.

For a cone opening about the

z

-axis, would you use cylindrical or spherical?

Either can work, but if the cone's boundary is a fixed angle from the axis, spherical makes it a constant

ϕ

; if it's described by

z

r

(like

z = r

), cylindrical is cleaner. Match the coordinate whose constant surfaces are the region's faces.

θ

ranges only over

[0, π]

instead of

[0, 2 π)

, what have you physically restricted?

You've kept only half the rotation around the

z

-axis — one side of a plane through that axis. Useful for half-cylinders or half-balls, but forgetting it silently halves your answer.

What happens to the tetrahedron volume

\frac{1}{6}

example if you set the innermost limit

z

to a constant instead of

1 - x - y

You'd integrate over a box, not the tetrahedron, getting the wrong region and wrong volume. The slanted face

x + y + z = 1

must appear as the

z

-upper-limit

1 - x - y

Can the density

f = ρ (x, y, z)

(mass density) ever legitimately be negative in

∭_{E} ρ d V

Not for physical mass — density is

\geq 0

, so mass is

\geq 0

. A negative "

f

" is fine mathematically (net charge, signed quantities), but then the result is a signed total, not a mass; see Center of mass and moments of inertia.

Active recall

Recall One-line self-test

Cover the answers and defend each in a full sentence.

Where does the $r$ in cylindrical $d V$ come from? ::: The arc-length $r d θ$ of a fan-shaped wedge that widens with distance from the axis.
Why $sin ϕ$ and not $sin θ$ ? ::: Because latitude circles shrink at the poles ( $ϕ = 0, π$ ), which $ϕ$ controls, not $θ$ .
Why must outer limits be constants? ::: Because inner variables are already integrated away, so only a number can remain outside.
When do you switch to spherical on sight? ::: When you see $x^{2} + y^{2} + z^{2}$ , since it becomes the single clean variable $ρ^{2}$ .

Connections

Triple integrals in Cartesian, cylindrical, spherical coordinates — the parent this page stress-tests.
Jacobian and change of variables — the source of every "extra factor" trap here.
Double integrals & polar coordinates — the 2D root of the cylindrical $r$ .
Spherical coordinates geometry — why $ϕ \in [0, π]$ and where $sin ϕ$ comes from geometrically.
Center of mass and moments of inertia — where signed and weighted integrands appear.