Math 314 Lecture #25
§15.8: Triple Integrals in Cylindrical Coordinates
Outcome A: Convert an equation from rectangular coordinates to cylindrical coordinates,
and vice versa.
The cylindrical coordinate system describes a point (x, y, z) in rectangular space in
terms of the triple (r, θ, z) where r and θ are the polar coordinates of the projection
(x, y, 0) and z is the (signed) distance from the projection (x, y, 0) to (x, y, z).
The transformation from cylindrical coordinates to rectangular coordinates is
x = r cos θ, y = r sin θ, z = z,
and the inverse transformation is
r2 = x2 + y 2 , tan θ = y/x, z = z.
We can transformation equations in x, y, and z to equations in r, θ, and z, and vice
versa.
Examples. (a) The hyperboloid of two sheets −x2 − y 2 + z 2 = 1 becomes
−r2 + z 2 = 1 or z 2 = 1 + r2 .
(b) The shifted paraboloid x2 + y 2 − z = 2y becomes
r2 − z = 2r sin θ or z = r2 − r sin θ = r(r − sin θ).
(c) The equation 2r2 + z 2 = 1 becomes the ellipsoid
2(x2 + y 2 ) + z 2 = 1.
(d) The equation r = 5 becomes the cylinder
x2 + y 2 = 25.
Outcome B: Describe a solid in cylindrical coordinates.
Cylindrical coordinates are useful in problems where there is a symmetry about the
z-axis.
Examples. (a) A solid lies between the cylinder x2 +y 2 = 1 and the sphere x2 +y 2 +z 2 =
4.
In cylindrical coordinates, these equations are
r2 = 1, r2 + z 2 = 4.
These surfaces intersect along the two curves given by
r2 = 1, z 2 = 3.
The cylindrical coordinate description of the solid is
√
√
{(r, θ, z) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, − 4 − r2 ≤ z ≤ 4 − r2 .
Here is a picture of this solid.
(b) What is the solid {(r, θ, z) : 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 2, r ≤ z ≤ 2}?
The solid lies between z = r and z = 2 over the first quadrant 0 ≤ θ ≤ π/2.
The surfaces z = r and z = 2 intersect along the curve r = 2, z = 2.
p
The surface z = r is the cone z = x2 + y 2 , and so the solid lies above this cone, below
the horizontal plane z = 2, all over that part of the disk of radius 2 in the first quadrant
in the xy-plane.
Here is a picture of this solid.
Outcome C: Evaluate a triple integral by converting it to cylindrical coordinates.
Conversion of a triple integral to cylindrical coordinates MAY simplify the iterated integration.
Let f (x, y, z) be continuous on the solid
E = {(x, y, z) : (x, y) ∈ D, u1 (x, y) ≤ z ≤ u2 (x, y)}
where
D = {(x, y) : α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)}.
The triple integral of f in cylindrical coordinates is
#
ZZZ
Z Z "Z
u2 (x,y)
f dV =
f (x, y, z) dz dA
E
D
Z
β
u1 (x,y)
Z
h2 (θ)
Z
u2 (r cos θ,r sin θ)
f (r cos θ, r sin θ, z) rdrdθ.
=
α
h1 (θ)
u1 (r cos θ,r sin θ)
Example. Let E be the solid enclosed by the two planes z = 0, z = x + y + 5 and
between the two cylinders x2 + y 2 = 4, x2 + y 2 = 9.
The bottom of this solid is z = 0, and the top of this surface is given by z = x + y + 5,
or in cylindrical coordinates,
z = r cos θ + r sin θ + 5 = r(cos θ + sin θ) + 5.
The “sides” of this solid are the cylinders, r2 = 4 and r2 = 9.
A cylindrical coordinate description of this solid is
E = {(r, θ, z) : 2 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 5 + r(cos θ + sin θ)}.
Here is a picture of this solid.
The triple integral of f (x, y, z) = x over E is
Z
2π
0
Z
Z
3
2
2π
Z
Z
0
3
=
Z0 2π Z2 3
=
Z0 2π Z2 3
5+r(cos θ+sin θ)
r2 cos θ dzdrdθ
2
z=5+r(cos θ+sin θ)
zr cos θ z=0
drdθ
(5 + r(cos θ + sin θ) r2 cos θ drdθ
2
5r cos θ + r3 (cos2 θ + cos θ sin θ) drdθ
0
2
r=3
Z 2π 3
5r
r4
2
=
cos θ + (cos θ + cos θ sin θ)
dθ
3
4
0
r=2
Z 2π 5
81 − 16
2
27 − 8 cos θ +
cos θ + cos θ sin θ dθ
=
3
4
0
Z 2π 65
95
2
cos θ + (cos θ + cos θ sin θ dθ
=
3
4
0
Z 2π 65 1 + cos 2θ
95
cos θ +
+ cos θ sin θ dθ
=
3
4
2
0
2π
95 sin θ 65 θ sin 2θ sin2 θ
=
+
+
+
3
4 2
4
2
0
65π
=
.
4
=