27 triple integrals in spherical and cylindrical coordinates

Cylindrical and Spherical Coordinates

The cylindrical coordinate system is the combination
of using polar coordinates for points in the xy–plane
with z as the 3rd
coordinate.

Example A. a. Plot the point (3,120o
, 4)
in cylindrical coordinate. Convert it to
rectangular coordinate.
with z as the 3rd
coordinate.

, 4)
3
120o
x
y
z
with z as the 3rd
coordinate.

, 4)
3
120o
4
x
y
z
(3, 120o
, 4)
with z as the 3rd
coordinate.

, 4)
3
120o
4
x
y
z
(3, 120o
, 4)
x = 3cos(120o
) = –3/2
y = 3sin(120o
) = √3
Hence the point is (–3/2, √3, 4)
with z as the 3rd
coordinate.

, 4)
3
120o
4
x
y
z
(3, 120o
, 4)
x = 3cos(120o
) = –3/2
y = 3sin(120o
) = √3
b. Convert (3, –3, 1) into to
cylindrical coordinate.
with z as the 3rd
coordinate.

, 4)
3
120o
4
x
y
z
(3, 120o
, 4)
x = 3cos(120o
) = –3/2
y = 3sin(120o
) = √3
θ = 315o
, r = √9 + 9 = √18
Hence the point is (√18, 315o
,
1) the cylindrical coordinate.
with z as the 3rd
coordinate.

, 4)
3
120o
4
x
y
z
(3, 120o
, 4)
x = 3cos(120o
) = –3/2
y = 3sin(120o
) = √3
(√18, 315o
, 0)
θ = 315o
, r = √9 + 9 = √18
,
1) the cylindrical coordinate. x
y
z
with z as the 3rd
coordinate.

, 4)
3
120o
4
x
y
z
(3, 120o
, 4)
x = 3cos(120o
) = –3/2
y = 3sin(120o
) = √3
(√18, 315o
, 0)
θ = 315o
, r = √9 + 9 = √18
,
1) the cylindrical coordinate. x
y
z
(√18, 315o
, 1) = (3, –3, 1)
1
with z as the 3rd
coordinate.

The constant equations
r = k describes the
cylinder of radius k, thus
the name "cylindrical
coordinate".
Example: Sketch r = 2

r = k describes the
coordinate".
2

r = k describes the
coordinate".
2
θ = k describes the
vertical half plane through
the origin, at the angle k
with x-axis. (r > 0)

r = k describes the
coordinate".
2
Example: Sketch θ =3π/4

r = k describes the
coordinate".
X
Y
2
Example: Sketch θ =3π/4
3π/4
Y

In spherical coordinate, a point in space is
represented by (ρ, θ, φ) where ρ gives the distance
from the origin to the point with θ and φ as shown in
the figure.
θ
X
Y
Z
φ ρ

the figure. We restrict ρ > 0, 2π > θ > 0, and
π > φ > 0.
θ
X
Y
Z
φ ρ

π > φ > 0.
θ
X
Y
Z
φ ρ
r = ρsin(φ)

π > φ > 0.
θ
X
Y
Z
φ ρ
r = ρsin(φ)
z = ρcos(φ)

π > φ > 0.
θ
X
Y
Z
The main conversion rule that
connects all these system is that
r = ρsin(φ). φ ρ
r = ρsin(φ)
z = ρcos(φ)

π > φ > 0.
θ
X
Y
Z
r = ρsin(φ).
Hence x = ρsin(φ)cos(θ),
φ ρ
r = ρsin(φ)
z = ρcos(φ)

π > φ > 0.
θ
X
Y
Z
r = ρsin(φ).
y = ρsin(φ)sin(θ),
φ ρ
r = ρsin(φ)
z = ρcos(φ)

π > φ > 0.
θ
X
Y
Z
r = ρsin(φ).
z = ρcos(φ)
φ ρ
r = ρsin(φ)
z = ρcos(φ)

π > φ > 0.
θ
X
Y
Z
r = ρsin(φ).
z = ρcos(φ)
ρ = √x2
+ y2
+ z2
, tan(φ) = r/z
φ ρ
r = ρsin(φ)
z = ρcos(φ)

Example: Plot the point (3, 2π/3, π/4) in cylindrical
coordinate. Convert it to the other systems.

2π/3
3
X
Y
Z
π/4

r = 3sin(π/4)
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ)
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
= (-3√2)/4
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
= (-3√2)/4
y = rsin(θ) = 3√2/2 * sin(2π/3)
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
= (-3√2)/4
y = rsin(θ) = 3√2/2 * sin(2π/3)
= (3√6)/4
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
= (-3√2)/4
y = rsin(θ) = 3√2/2 * sin(2π/3)
= (3√6)/4
z = 3cos(π/4) = (3√3)/2 2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
= (-3√2)/4
y = rsin(θ) = 3√2/2 * sin(2π/3)
= (3√6)/4
z = 3cos(π/4) = (3√3)/2
Hence the point is:
((-3√2)/4, (3√6)/4, (3√3)/2)
in the rectangular coordinate;
2π/3
3
X
Y
Z
π/4

r = 3sin(π/4) = (3√2)/2
x = rcos(θ) = (3√2)/2 * cos(2π/3)
= (-3√2)/4
y = rsin(θ) = 3√2/2 * sin(2π/3)
= (3√6)/4
z = 3cos(π/4) = (3√3)/2
Hence the point is:
((-3√2)/4, (3√6)/4, (3√3)/2)
in the rectangular coordinate;
((3√2)/2, 2π/3, (3√3)/2) in
the cylindrical coordinate.
2π/3
3
X
Y
Z
π/4

ρ = k describe the sphere
of radius k, thus the name
"spherical coordinate".
Example: Sketch ρ = 2

ρ = k describe the sphere
of radius k, thus the name
"spherical coordinate".
Example: Sketch ρ = 2 Example: Sketch φ = π/2
φ = π/2
φ = k describe the cone of
of angle φ with the z axis.

Example: Convert the equation z = √k2
– x2
– y2
into
the polar equation and the spherical equation.

– x2
– y2
into
k

– x2
– y2
into
Since r = x2
+ y2
,
k

– x2
– y2
into
Since r = x2
+ y2
, the polar
equation is z = √k2
– r2
k

– x2
– y2
into
Since r = x2
+ y2
, the polar
equation is z = √k2
– r2
For spherical equation, we get
with 0 < φ < π/2
ρ = k
k

Example: Convert the equation z = x2
+ y2
into the
polar equation and the spherical equation.

+ y2
into the
Since r = √x2
+ y2
, the polar
equation is z = r2

+ y2
into the
Since r = √x2
+ y2
, the polar
equation is z = r2
r
z = r2

+ y2
into the
Since r = √x2
+ y2
, the polar
equation is z = r2
For spherical equation, set
z = ρcos(φ)r = ρsin(φ),
r
z = r2
ρ
φ

+ y2
into the
Since r = √x2
+ y2
, the polar
equation is z = r2
We get
= (ρsin(φ))2ρcos(φ)
r
z = r2
ρ
φ

+ y2
into the
Since r = √x2
+ y2
, the polar
equation is z = r2
We get
= (ρsin(φ))2ρcos(φ)
= ρ*sin2
(φ)cos(φ)
or r
z = r2
ρ
φ

Triple Integrals in Cylindrical and Spherical Coordinates

A solids V may be represented easier in cylindrical
or sphereical format than in rectangular
coordinates.

coordinates. Thus a triple integral of a function with
domain V might be easier to compute if we utilize
cylindrical or spherical coordinates.

Let w = h(r, θ, z ) be a function over
V in cylindrical coordinates.
Cylindrical Coordinates

We partion V by small increments
of Δr, Δθ, and Δz.

of Δr, Δθ, and Δz.
Δθ
X
Y
Z
Δr
Δr
Δz

of Δr, Δθ, and Δz. A typical
ΔV=ΔrxΔθxΔ is a cylindrical wedge
whose volume depends on r, i.e
its distance to the z-axis.
Δθ
X
Y
Z
Δr
Δr
Δz

In particular, let r*
be the
distance to the center of
ΔrxΔθ,
Δθ
X
Y
Z
Δr
Δz
r*

be the
ΔrxΔθ, then volume of
ΔV = r*
ΔrΔθΔz since the
base area is r*ΔrΔθ. Δθ
X
Y
Z
Δr
Δz
r*

be the
ΔV = r*
X
Y
Z
Δr
Δz
r*
Let (ri, θi, zi) is the center point of each wedge in the partition,
(ri, θi, zi)

be the
ΔV = r*
X
Y
Z
Δr
Δz
r*
= lim Σh(ri, θi, zi) ri ΔrΔθΔz,Δr,Δθ,Δz0
then h(r, θ, z) dV∫ ∫∫v
(ri, θi, zi)

be the
ΔV = r*
X
Y
Z
Δr
Δz
r*
which may be reformulated as an integral over drdθdz
depending on the description of V in polar coordinates.
(ri, θi, zi)

be the
ΔV = r*
X
Y
Z
Δr
Δz
r*
which may be reformulated as an integral over drdθdz
depending on the description of V in polar coordinates.
Usually V is given as {D; g(r, θ) < z < f(r, θ)} where D is the
polar representation of the base of V in the xy-plane, with
g(r, θ) and f(r, θ) form the floor and ceiling of V.
(ri, θi, zi)

X
Y
Z
π/6
π/6
3
5
7
V
Example: Write V using
cylindrical coordinates.

X
Y
Z
π/6
π/6
3
5
7
V
In cylindrical coordinate, V is
{π/6 < θ < π/3;

X
Y
Z
π/6
π/6
3
5
7
V
{π/6 < θ < π/3; 5 < r < 7;

X
Y
Z
π/6
π/6
3
5
7
V
{π/6 < θ < π/3; 5 < r < 7; 0 < z < 3}
or in any order of r, θ, and z.

X
Y
Z
π/6
π/6
3
5
7
V
{π/6 < θ < π/3; 5 < r < 7; 0 < z < 3}
z=√25–x2
–y2
3
3
V
D
1
5

X
Y
Z
π/6
π/6
3
5
7
V
{π/6 < θ < π/3; 5 < r < 7; 0 < z < 3}
The projection D of V onto the
xy-coordinate is
{0 < θ < 2π; 0 < r < 3}.
z=√25–x2
–y2
3
3
V
D
1
5

X
Y
Z
π/6
π/6
3
5
7
V
{π/6 < θ < π/3; 5 < r < 7; 0 < z < 3}
xy-coordinate is
{0 < θ < 2π; 0 < r < 3}.
z=√25–x2
–y2
3
3
V
D
1
5
z = √25 – x2
– y2
is z = √25 – r2

X
Y
Z
π/6
π/6
3
5
7
V
{π/6 < θ < π/3; 5 < r < 7; 0 < z < 3}
xy-coordinate is
{0 < θ < 2π; 0 < r < 3}.
z=√25–x2
–y2
3
3
V
D
1
5
z = √25 – x2
– y2
is z = √25 – r2
Hence V is {0 < θ < 2π; 0 < r < 3; 1 < z < √25 – r2
}.

=then h(r, θ, z) dV∫ ∫∫v
Theorem: Given a solid V in cylindrical coordinate as
{A < θ < B; g(θ) < r < f(θ); G(r, θ) < z < F(r, θ) }
h(r, θ, z) r dz dr dθ∫ ∫∫
B
θ=A r=g(θ)
f(θ)
z=G(r,θ)
F(r,θ)
In paricular if h(r, θ, z) = 1, we obtain the volume of V.

B
θ=A r=g(θ)
f(θ)
z=G(r,θ)
F(r,θ)
Example: Find the volume of V
using cylindrical coordinates.
z=√25–x2
–y2
3
3
V
D
1
5

B
θ=A r=g(θ)
f(θ)
z=G(r,θ)
F(r,θ)
z=√25–x2
–y2
3
3
V
D
1
5
Since V is
{0 < θ < 2π; 0 < r < 3; 1 < z < √25 – r2
}.

B
θ=A r=g(θ)
f(θ)
z=G(r,θ)
F(r,θ)
z=√25–x2
–y2
3
3
V
D
1
5
Since V is
{0 < θ < 2π; 0 < r < 3; 1 < z < √25 – r2
}.
Therefore it's volume is
1* r dz dr dθ∫ ∫∫
2π
θ=0 r=0
3
z=1
√25 – r2

2π
θ=0 r=0
3
z=1
√25 – r2

2π
θ=0 r=0
3
z=1
√25 – r2
= rz | dr dθ∫ ∫
2π
θ=0 r=0
3
z=1
√25 – r2

2π
θ=0 r=0
3
z=1
√25 – r2
2π
θ=0 r=0
3
z=1
√25 – r2
= r(25 – r2
)1/2
– r dr dθ∫ ∫
2π
θ=0 r=0
3

2π
θ=0 r=0
3
z=1
√25 – r2
2π
θ=0 r=0
3
z=1
√25 – r2
= r(25 – r2
)1/2
– r dr dθ∫ ∫
2π
θ=0 r=0
3
= – (25 – r2
)3/2
– ½ r2
| dθ∫
2π
θ=0 r=0
3
3
1

2π
θ=0 r=0
3
z=1
√25 – r2
2π
θ=0 r=0
3
z=1
√25 – r2
= r(25 – r2
)1/2
– r dr dθ∫ ∫
2π
θ=0 r=0
3
= – (25 – r2
)3/2
– ½ r2
| dθ∫
2π
θ=0 r=0
3
3
1
= dθ∫
2π
θ=0
6
95
=
3
95π

Let w = h(ρ, θ, φ ) be a function
over V in sphereical coordinates.
Spherical Coordinates

of Δρ, Δθ, and Δφ.

of Δρ, Δθ, and Δφ. A typical
ΔV=ΔρxΔθxΔφ is a chunk of a
spherical wedge.
Δθ
X
Y
Z
Δφ
Δρ
Δφ

spherical wedge.
Δθ
X
Y
Z
Δφ
The volume of ΔV depends on ρ-the distance to
the origin, and φ-the pitched angel.
Δρ
Δφ

spherical wedge.
Δθ
X
Y
Z
Δφ
The larger the ρ is, the larger the volume of ΔV.
The closer φ is to π/2, the larger the volume of ΔV is.
Δρ
Δφ

spherical wedge.
Δθ
X
Y
Z
Δφ
The larger the ρ is, the larger the volume of ΔV.
The closer φ is to π/2, the larger the volume of ΔV is.
Δρ
Δφ
In fact, the volume of ΔV is (ρ*
)2
sin(φ*
)ΔρΔθΔφ where
(ρ*
, θ*
, φ*
) is the center of ΔV.

Δρ,Δθ,Δφ0
Then∫ ∫∫v
h(ρ, θ, φ )dV = lim Σ h(ρ*, θ*, φ*) ΔV

Δρ,Δθ,Δφ0
Then∫ ∫∫v
h(ρ, θ, φ )dV = lim Σ h(ρ*, θ*, φ*)
(ρ*)2
sin(φ*) ΔρΔθΔφ
ΔV
Δρ,Δθ,Δφ0
= lim Σ h(ρ*, θ*, φ*)

Δρ,Δθ,Δφ0
Then∫ ∫∫v
(ρ*)2
ΔV
Δρ,Δθ,Δφ0
= lim Σ h(ρ*, θ*, φ*)
ρ2
sin(φ)=∫ ∫∫
appropriate limits
for integration
h(ρ, θ, φ ) dρ dθ dφ

Δρ,Δθ,Δφ0
Then∫ ∫∫v
Its easier to express solids related to the spheres in sperical
coordinates:
(ρ*)2
ΔV
Δρ,Δθ,Δφ0
= lim Σ h(ρ*, θ*, φ*)
ρ2
sin(φ)=∫ ∫∫
appropriate limits
for integration
h(ρ, θ, φ ) dρ dθ dφ

Δρ,Δθ,Δφ0
Then∫ ∫∫v
Its easier to express solids related to the spheres in sperical
coordinates:
(ρ*)2
ΔV
Δρ,Δθ,Δφ0
= lim Σ h(ρ*, θ*, φ*)
ρ2
sin(φ)=∫ ∫∫
appropriate limits
for integration
h(ρ, θ, φ )
{0 < ρ < R; 0 < θ < 2π; 0 < φ < π }
The sphere of radius R:
R
dρ dθ dφ

If we varies ρ, we get the spherical shell:

{r < ρ < R; 0 < θ < 2π; 0 < φ < π }
Rr

{r < ρ < R; 0 < θ < 2π; 0 < φ < π }
Rr
If we varies θ, we get the spherical wedge:

{r < ρ < R; 0 < θ < 2π; 0 < φ < π }
Rr
{0 < ρ < R; A < θ < B; 0 < φ < π }
If we varies θ, we get the spherical wedge:
θ=A
θ=B

If we varies φ, we get the spherical cone:

{0 < ρ < R; 0 < θ < 2π; 0 < φ < C }
R
φ=C

{0 < ρ < R; 0 < θ < 2π; 0 < φ < C }
R
If we varies all tree variables, we get a chunk of the
spherical wedge:
φ=C

{0 < ρ < R; 0 < θ < 2π; 0 < φ < C }
R
{r < ρ < R; A < θ < B; C < φ < D }
If we varies all tree variables, we get a chunk of the
spherical wedge:
θ=A
θ=B
φ=C
φ=C
ρ=r φ=D
ρ=R

Example: Find∫ ∫∫v
z dV
where V is the solid bounded above
by z = √1 – x2
– y2
and below by
z = √3x2
+ 3y2

z dV
by z = √1 – x2
– y2
and below by
z = √3x2
+ 3y2 1
x
y

z dV
by z = √1 – x2
– y2
and below by
z = √3x2
+ 3y2
Put V in spherical coordinates.
The pitched angle φ is π/6 because
the border of the cone in the
yz-plane is z = √3y.
1
x
y

z dV
by z = √1 – x2
– y2
and below by
z = √3x2
+ 3y2
1
φ=π/6
x
y

z dV
by z = √1 – x2
– y2
and below by
z = √3x2
+ 3y2
So V = {0 < ρ < 1;0 < θ < 2π ; 0 < φ < π/6}.
The integrand z = ρcos(φ), hence the integral is
1
φ=π/6
x
y

z dV
by z = √1 – x2
– y2
and below by
z = √3x2
+ 3y2
So V = {0 < ρ < 1;0 < θ < 2π ; 0 < φ < π/6}.
The integrand z = ρcos(φ), hence the integral is
ρ2
sin(φ)=∫ ∫∫ ρ=0
ρ2
cos(φ)
π/6 2π 1
θ=0φ=0
1
φ=π/6
x
y
dρ dθ dφ

ρ2
sin(φ)∫ ∫∫ ρ=0
ρ2
cos(φ)
π/6 2π 1
θ=0φ=0
dρ dθ dφ

ρ2
sin(φ)∫ ∫∫ ρ=0
ρ2
cos(φ)
π/6 2π 1
θ=0φ=0
sin(φ)= ∫ ∫ dθ dφcos(φ)
π/6 2π
θ=0φ=0
1
5
dρ dθ dφ

ρ2
sin(φ)∫ ∫∫ ρ=0
ρ2
cos(φ)
π/6 2π 1
θ=0φ=0
π/6 2π
θ=0φ=0
1
5
sin(φ)= ∫ dφcos(φ)
π/6
φ=0
2π
5
dρ dθ dφ

ρ2
sin(φ)∫ ∫∫ ρ=0
ρ2
cos(φ)
π/6 2π 1
θ=0φ=0
π/6 2π
θ=0φ=0
1
5
sin(φ)= ∫ dφcos(φ)
π/6
φ=0
2π
5
sin2
(φ) |=
π/6
φ=0
π
5
=
π
20
dρ dθ dφ

A note on calculation of double and triple integrals:
If the domnain of a double integral is a rectangle
{a < x < b; c < y < d} and the integrand is of the
form = f(x)g(y), then

f(x) g(y)∫ ∫ dy dx =
y=cx=a
b d
g(y)∫ ∫ dy
y=cx=a
b d
f(x)dx *

f(x) g(y)∫ ∫ dy dx =
y=cx=a
b d
g(y)∫ ∫ dy
y=cx=a
b d
f(x)dx *
Similarly for triple integrals, we've:
f(x) g(y) h(z)∫ ∫ dy dx dz =
y=cx=a
b d
g(y)∫ ∫ dy *
y=cx=a
b d
f(x) dx*∫z=e
f
h(z)∫ dz
z=e
f
Exercise: Do the integral in the last example via this observation.

27 triple integrals in spherical and cylindrical coordinates

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27 triple integrals in spherical and cylindrical coordinates