II.f Triple Integrals in Cylindrical and Spherical Coordinates
We have already seen the advantage of changing to polar coordinates in some
double integral problems. The same situation happens for triple integrals. For this
case we introduce two coordinate systems besides our friend (x, y, z): the first is
(r, θ, z) with (r, θ) as given before; the second is (ρ, ϕ, θ) which we define shortly.
We remark that there are many other coordinate systems which you may encounter
in applications, but only the ones given above will be developed in this course.
Since (r, θ, z) is essentially familiar to us, we merely note that dV in this case
is dV = (r dr dθ) dz. You should be aware that there are two other versions of this
system: (r, θ, y) and (r, θ, x). For (r, θ, y) we set
z
z = r sin θ
r
θ
x
x= r cos θ
while for (r, θ, x) we have
z
z= r sin θ
r
y= r cos θ.
θ
y
We select one of the three possibilities based, once again, on how the boundary of
the solid is given. Please see the examples below.
148
We now pass to the spherical system: (ρ, ϕ, θ). This is defined as follows:
p

x2 + y 2 + z 2
ρ
=





z
z

cos ϕ = = p
ρ
x2 + y 2 + z 2





y

tan θ =
x


 z = ρ cos ϕ
y = ρ sin ϕ sin θ


x = ρ sin ϕ cos θ.
z
ρ cos φ
(x , y , z)
θ
ρ
φ
ρs
in φ
y
x
These formulas enable us to move between the (x, y, z) system and the (ρ, ϕ, θ)
system. We note the following: (a) ρ measures the distance of a point to the
origin, and is thus nonnegative (unlike r); (b) ϕ corresponds to “latitude”; (c) θ
corresponds to “longitude.” Note that ϕ varies between 0 and π (not 2π) and ϕ = 0
corresponds to the positive z-axis. On the other hand, θ varies between 0 and 2π,
just like before. To become a little more familiar with (ρ, ϕ, θ) we note:
Example 1.
The sphere x2 + y 2 + z 2 = a2 becomes
ρ2 sin2 ϕ cos2 θ + ρ2 sin2 ϕ sin2 θ + ρ2 cos2 ϕ = a2
or ρ = a.
149
z
ρ=a
y
x
Example 2.
p
x2 + y 2 becomes
The cone z =
q
q
2
2
2
2
2
2
ρ cos ϕ = ρ sin ϕ cos θ + ρ sin ϕ sin θ = ρ2 sin2 ϕ = |ρ sin ϕ|
or tan ϕ = 1, i.e., ϕ = π/4.
z
π/ 4
φ =π/ 4
y
x
Example 3.
The plane: x − y = 0 becomes
ρ sin ϕ cos θ = ρ sin ϕ sin θ
or tan θ = 1, i.e., θ = π/4 or θ = 5π/4.
z
θ = 5π/4
θ=π/4
y
x
150
=
↑
(sin ϕ≥0!)
ρ sin ϕ
What these three example show is that the surfaces ρ = constant are spheres;
the surfaces ϕ = constant are cones; the surfaces θ = constant are 1/2 planes. This
coordinate system should always be considered for triple integrals where f (x, y, z)
becomes simpler when written in spherical coordinates and/or the boundary of the
solid involves (some) cones and/or spheres and/or planes.
We now consider the volume element dV in terms of (ρ, ϕ, θ). Suppose we
increase ρ by dρ, ϕ by dϕ and θ by dθ. We obtain a volume dV as shown in the
figure.
z
dV
ρ sin φ
dρ
ρ sin φ d θ
dθ
φ
ρdφ
dφ
y
x
Just as had been the case for dA in polar coordinates, in the limit we can treat
dV as if it were a cube (this can be formally proved in theoretical mathematics
courses). We then get
dV = (dρ)(ρ dϕ)(ρ sin ϕ dθ) = ρ2 sin ϕ dρ dϕ dθ.
Observe two facts:
(1) the units for dV are correct (length3 ),
(2) the presence of the term “sin ϕ” in (ρ sin ϕ dθ). This is due to the fact that as
θ varies, the “arm of rotation” is ρ sin ϕ and not ρ. (Think of what happens
if ϕ is almost zero!)
151
We now pass to examples.
Example 4.
Find the mass of the solid bounded by z = x2 + y 2 − 4 and z = 0 if
the density f (x, y, z) = 1 + x2 + y 2 .
Answer.
We note that this problem begs for cylindrical coordinates. Indeed, z = 0
corresponds to x2 + y 2 = 4 (i.e., r = 2) and f = 1 + r 2 !
z
x 2 + y2 = 4
y
x
z = x 2 + y2 - 4
We can change to cylindrical coordinates immediately or half way through! Suppose
we start with (x, y, z). Then we get
Mass of cube = (1 + x2 + y 2 ) dx dy dz
Z z=0 (top)
Mass of column =
(1 + x2 + y 2 ) dz dx dy
z=x2 +y 2 −4 (bottom)
y
x 2 + y2 = 4
x
152
Mass of slice =
Z √4−y 2
x=−
Mass =
Z
√
4−y 2
1
y=−1
Z
z=0
(1 + x2 + y 2 ) dz dx dy
z=x2 +y 2 −4
Z √4−y 2
x=−
√
4−y 2
Z
z=0
(1 + x2 + y 2 ) dz dx dy.
z=x2 +y 2 −4
Clearly evaluating this integral is going to be a lot of work! Instead, we switch to
cylindrical coordinates.
y
z
y
x
x
Mass of “cube” = (1 + x2 + y 2 ) dz r dr dθ = (1 + r 2 )r dz dr dθ
Z 0
Mass of column =
(1 + r 2 )r dz dr dθ
z=r 2 −4
Mass of slice =
Z
Mass =
Z
=
Z
2
r=0
2π
θ=0
2π
θ=0
=−
Z
=−
Z
Z
0
(1 + r 2 )r dz dr dθ
z=r 2 −4
Z
Z
2
r=0
2
r=0
2π
θ=0
2π
θ=0
Z
Z
0
(r + r 3 )dz dr dθ
z=r 2 −4
−(r + r 3 )(r 2 − 4)dr dθ
2
r=0
(−4r + r 5 − 3r 3 ) dr dθ
r6
3r 4
−2r +
−
6
4
2
153
2
0
dθ
64
64
+ 12 = 2π 20 −
= 2π 8 −
6
6
Example 5. Find the z coordinate of the center of mass of the solid consisting
p
of the part of the hemisphere z = 4 − x2 − y 2 inside the cylinder x2 + y 2 = 2x if
the density ρ = 1.
Answer.
Again we try using cylindrical coordinates, this time from the start. Note
that x2 + y 2 = 2x is not centered at the origin.
z
z = 4 - x 2- y 2
y
x
x 2 + y2 = 2x
We have, as before,
Mass of “cube” = ρ dV = 1(r dr dθ) dz
Mass of column =
Z
√
4−r2
r dz dr dθ.
z=0
Looking down we see the picture given below.
154
r=2
r = 2 cos θ
θ=π/2
θ=−π /2
Note that x2 + y 2 = 2x becomes r = 2 cos θ, and thus
Z
Mass =
π
2
θ=− π
2
Z
2 cos θ
r=0
Z
√
4−r2
r dz dr dθ.
z=0
(Note that we could also use symmetry here. We do instead the given integral to
practice.) So,
Mass =
=
Z
Z
π
2
−π
2
π
2
−π
2
1
=
3
Z
1
=
3
Z
Z
2 cos θ
r
r=0
"
π
2
p
4 − r 2 dr dθ
2 cos θ #
(4 − r 2 )3/2 −
dθ
3
r=0
−π
2
π
2
−π
2
(u = 4 − r 2 )
[8 − (4 − 4 cos2 θ)3/2 ] dθ
h
2
8 − (4 sin θ)
3/2
i
dθ.
Attention: It would seem reasonable to write (4 sin2 θ)3/2 = 8 sin3 θ but this is
√
WRONG! Actually, (4 sin2 θ)3/2 = ( 4 sin2 θ)3 = 8| sin θ|3 and, if −π/2 ≤ θ ≤ 0,
| sin θ| = − sin θ! So, we split the integral:
1
Mass =
3
Z
π
2
0
3
8(1 − sin θ) dθ +
155
Z
0
−π
2
8(1 + sin θ) dθ .
3
But,
Z
Z
π
2
0
0
−π
2
and so
π
cos3 θ 2
2
= ,
sin θ dθ =
(1 − cos θ) sin θ dθ = − cos θ +
3 0
3
0
0
2
cos3 θ 3
=−
sin θ dθ = − cos θ +
3 −π
3
Z
3
π
2
2
2
8 π 2
8 π 2
16 π 2
Mass =
+
=
.
−
−
−
3 2
3
3 2
3
3 2
3
Now for Mxy
Mxy =
=
=
Z
Z
Z
π
2
θ=− π
2
π
2
θ=− π
2
π
2
−π
2
π
2
Z
2 cos θ
r=0
Z
r4
r −
8
2
√
4−r2
z=0
2 cos θ
dθ =
r=0
Z
π
2
−π
2
π
2
2 cos θ
r
(4 − r 2 ) dr dθ
π
2
θ=− 2 r=0
16 cos4 θ
2
4 cos θ −
dθ
8
zr dz dr dθ =
Z
Z
2(1 + cos 2θ) − (1 + cos 2θ)2 dθ
1 + cos 4θ
dθ
=
2(1 + cos 2θ) − 1 + 2 cos 2θ +
2
−π
2
3
π
= 2−
·π = .
2
2
Z
So
z=
π
3
·
.
π
2 16[ 2 − 23 ]
We now consider some examples using spherical coordinates.
Example 6.
Find the mass of the hollow region bounded by the spheres x2 + y 2 +
z 2 = 4 and x2 + y 2 + z 2 = 1 if the density is directly proportional to the distance
from the origin.
156
Answer.
z
x 2 + y 2 + z2 = 1 ( or ρ = 1)
x 2 + y2 + z2 = 4
( or ρ = 2)
y
x
Now
dV = ρ2 sin ϕ dρ dϕ dθ
k
k
density = p
= .
ρ
x2 + y 2 + z 2
So we get
2π
π
2
k 2
ρ sin ϕ dρ dϕ dθ
θ=0 ϕ=0 ρ=1 ρ
2 Z 2π Z π k
2
dϕ dθ
=
(sin ϕ) ρ θ=0 ϕ=0 2
ρ=1
Z 2π Z π
k
=
sin ϕ[4 − 1] dϕ dθ
θ=0 ϕ=0 2
Z 2π
π
3
3
= k
− cos ϕ 0 dθ = k · 2 · 2π = 6kπ.
2 0
2
Mass M =
Z
Z
Z
Can you imagine how hard the integrals would be if we tried to do this problem
using (x, y, z)!
157
Example 7.
Evaluate by changing to spherical coordinates:
Z 5 Z √25−x2 Z √50−x2 −y 2 p
x2 + y 2 + z 2 dz dy dx.
√
x=0
Answer.
y=0
z=
x2 +y 2
We start by sketching the region. Again, no work of art is needed, but
the picture should be good enough to determine the limits of integration. We first
observe that the first integral involves z, so the columns run up and down. So we
look down on the solid, and we note that x goes from 0 to 5 and for any x, y goes
√
from zero to y = 25 − x2 , i.e., x2 + y 2 = 25. It follows that if we look down on
the solid, we see a quarter of a circle.
y
x
p
Now for a given (x, y) in the quarter circle, z goes between z = x2 + y 2 (i.e., a
p
cone) and z = 50 − x2 − y 2 (i.e., x2 + y 2 + z 2 = 50, and thus a sphere). Note that
p
the cone z = x2 + y 2 and the sphere x2 +y 2 +z 2 = 50 meet at x2 +y 2 +x2 +y 2 = 50,
√
i.e., x2 +y 2 = 25 (and z = 25 = 5 but we shall not need this). So our solid consists
of the points inside the cone and sphere and over the quarter circle.
158
z
y
x
Now
(
z=
p
x2 + y 2
becomes
x2 + y 2 + z 2 = 50 becomes
tan ϕ = 1
√
ρ = 50.
or ϕ =
π
4
So
p
x2 + y 2 + z 2 dV
(ρ)ρ2 sin ϕ dρ dϕ dθ
=⇒
and our integral is
Z
π
2
θ=0
Z
π
4
ϕ=0
Z
√
50
3
ρ sin ϕ dρ dϕ dθ =
ρ=0
=
Z
Z
π
2
θ=0
π
4
0
Z
π
4
ϕ=0
(50)2
sin ϕ dϕ dθ
4
"
√ #
π4
(50)2 2
(50)2
π
· 1−
− cos ϕ 0 dθ =
· .
4
4
2
4
Example 8. Find the z coordinate of the centroid of a cone of height h, base
radius b, positioned so that the vertex is at (0, 0, 0) and the axis is the positive
z-axis.
159
Answer.
z
b
h
y
x
Note that the equation of the cone is
z=
hp 2
x + y2.
b
We calculate the volume first. We have a choice of approaches and we use spherical
coordinates to practice. The top of the cone is z = h, i.e. ρ cos ϕ = h or ρ = h sec ϕ,
p
while z = (h/b) x2 + y 2 becomes cos ϕ = (h/b) sin ϕ =⇒ tan ϕ = b/h. We then
have
V =
=
Z
2π
θ=0
1
3
Z
h3
=
3
0
Z
b
tan−1 ( h
)
ϕ=0
2π Z
Z
2π
0
tan
−1
Z
h sec ϕ
ρ2 sin ϕ dρ dϕ dθ
ρ=0
b
(h
)
ϕ=0
cos−2 ϕ
2
h3
sin ϕ dϕ dθ
cos3 ϕ
tan−1 ( hb )
dθ.
ϕ=0
Now sec2 ϕ = 1 + tan2 ϕ and so
sec
2
tan
−1
b2
b
=1+ 2
h
h
and we get
h3
V =
3·2
Z
2π
0
b2
1+ 2
h
h3 b 2
1
− 1 dθ =
· 2 · 2π = πb2 h,
6 h
3
160
just like the known formula! Now
Mxy =
=
Z
Z
2π
θ=0
2π
θ=0
2π
Z
Z
b
tan−1 ( h
)
ϕ=0
tan
−1
Z
h sec ϕ
(ρ cos ϕ)ρ2 sin ϕ dρ dϕ dθ
ρ=0
b
(h
)
(cos ϕ)(sin ϕ)
ϕ=0
1
h4
dϕ dθ
4 cos4 ϕ
b
)
tan−1 ( h
h4 sin ϕ
dϕ dθ
4 cos3 θ
θ=0 ϕ=0
tan−1 ( hb )
Z 2π 4 h
h4 b 2
π
1
dθ
=
· 2 · 2π = h2 b2
=
2
4 · 2 cos ϕ ϕ=0
8 h
4
0
=
and so
Z
Z
z=
π h2 b 2 3
3
= h.
2
4 πb h
4
To practice, let us do the same problem using cylindrical coordinates. We have
V =
=
=
Z
Z
2π
θ=0
2π
θ=0
Z
Z
b
r=0
b
r=0
Z
h
dz r dr dθ
z= h
br
hr
h−
b
r dr dθ =
πhb2
hb2
· 2π =
,
6
3
Z
2π
θ=0
hb2
h b3
− ·
2
b 3
dθ
1
r dr dθ
2
while
Mxy =
=
Z
Z
2π
θ=0
2π
θ=0
Z
b
r=0
Z
h
z dz r dr dθ =
z= h
br
1 h2 b 2
h2 b
− 2
2
2
b 4
4
dθ =
Exactly the same as before!
161
Z
2π
θ=0
Z
b
r=0
h2 r 2
h − 2
b
2
1 2 2
πh2 b2
h b · 2π =
.
8
4
Further Exercises:
1) Find the volume of the solid bounded by the cone x =
plane x = 1.
p
y 2 + z 2 and the
2) Find the volume of the solid bounded by the sphere x2 + y 2 + z 2 = 1 and the
cylinder x2 + y 2 = 2y.
3) Find the radius of gyration of the cylinder bounded by x2 + y 2 = 1, z = 1,
z = −1 if the density is constant.
4) Find the center of mass of the hemisphere bounded by the surfaces z =
p
p
− 1 − x2 − y 2 , z = 0 if the density is 1 + x2 + y 2 .
5) Calculate the volume of the ellipsoid x2 +
z2
y2
+
= 1.
4
9
6) Find the moment of inertia with respect to the xy plane of the solid bounded
p
by: the hemisphere z =
9 − x2 − y 2 , the plane z = 0 and the surface
x2 + y 2 = 3x if the density is constant.
7) Calculate the centroid of the solid bounded by z = 1 −
p
x2 + y 2 and z = −3.
8) Calculate the centroid of the solid bounded by the cone z =
the planes z = 1, z = 2.
p
x2 + y 2 and
9) Calculate the volume inside the cylinder x2 + y 2 = 1 and outside the cone
z 2 = x2 + y 2 .
10) Find the mass of the solid bounded by the surfaces: ρ = sec ϕ, ϕ = π/4,
θ = π/2, θ = π if the density is constant.
162