Spherical Coordiantes
Multiple Integrals
Advanced Calculus Lecture 4 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 16, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Spherical coordinates In the system of spherical coordinates a point P = [x, y, z] is represented by x y z
= ρ sin φ cos θ, = ρ sin φ sin θ, = ρ cos φ,
0≤φ≤π 0 ≤ θ ≤ 2π . ρ≥0
Note that ρ2 = x 2 + y 2 + z 2 = r 2 + z 2 . The r coordinate in cylindrical coordinates is related to ρ and φ by q r = x 2 + y 2 = ρ sin φ . also
r tan φ = = z
p
x2 + y2 z
Dr. Lahcen Laayouni
and Advanced Calculus
tan θ =
y . x
Multiple Integrals
Spherical coordinates
Triple integrals
The volume element in spherical coordinates The volume element in spherical coordinates is given by dV = ρ2 sin φd ρd φd θ . If f is a continuous function on the bounded region R , then ZZZ ZZZ f (x, y, z)dV = g(ρ, φ, θ)ρ2 sin φd ρd φd θ , R
S
where g(ρ, φ, θ) = f (x, y, z) and S is the region R expressed in spherical coordinates. Remark Spherical coordinates are suited to problems involving spherical symmetry and, in particular, to regions bounded by spheres centered at the origin, circular cones with axes along the z-axis, and vertical planes containing the z-axis. Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example Evaluate
I=
ZZZ
(1 + x 2 + y 2 )dV ,
R
3.0 2.5 2.0
where R is the region in the first octant bounded by two spheres of radius 2 and 3 respectively.
z1.5 1.0 3.0 0.5
2.0
1.5 0.0 0.0 0.0
0.5 0.5
1.0
1.0
2.5
y
1.5 x
2.0
2.5
3.0
In spherical coordinates, the region R is given by the inequalities 2 ≤ ρ ≤ 3 , 0 ≤ φ ≤ π/2 , 0 ≤ θ ≤ π/2 , and the integrand becomes 1 + x 2 + y 2 = 1 + r 2 = 1 + ρ2 sin2 φ . Then Z π/2 Z π/2 Z 3 I = dθ dφ (1 + ρ2 sin2 φ)ρ2 sin φd ρ 0
0
Dr. Lahcen Laayouni
2
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example Thus I =
Z
π/2
dθ
0
= = = =
Z
π/2
dφ
0
Z
3
(ρ2 sin φ + ρ4 sin3 φ)d ρ
2
ρ=3 ρ3 ρ5 3 dθ dφ sin φ + sin φ 3 5 0 0 ρ=2 Z π/2 Z π/2 19 211 dθ sin φ + sin3 φ d φ 3 5 0 0 φ=π/2 Z π/2 517 211 2 dθ − cos φ − sin φ cos φ 15 15 0 φ=0 Z π/2 517 517 dθ = π. 15 0 30 Z
π/2
Z
π/2
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example Evaluate I =
ZZZ q
x 2 + y 2 + z 2 dV ,
E
where E
1.0
is the solid bounded by
0.75 z
z=1,
z 2 = x 2 + y 2 , above the
−1.0
xy -plane.
0.5
0.25 x y −0.5 −0.5 0.0 0.0
−1.0
0.0
0.5
0.5 1.0
1.0
In spherical coordinates the equation of the cone becomes (ρ cos φ)2 = (ρ sin φ cos θ)2 + (ρ sin φ sin θ)2 , which reduces to cos2 φ = sin2 φ(cos2 θ + sin2 θ) = sin2 φ . or cos φ = sin φ , from which we find that tan φ = 1 .Thus the cone is simply defined by 0 ≤ φ ≤ π/4 . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example The plane is described by ρ cos φ = 1 , or ρ = sec φ , thus Z 2π Z π/4 Z sec φ I = dθ dφ ρρ2 sin φd ρ 0
= = =
Z Z
0
2π
dθ 0
dθ 0
= −
π/4
Z
π/4
sin φd φ
0
2π
1 4
0
Z
2π
dθ
0
2π 4
Z
1
sec φ
ρ3 d ρ
0
sin φd φ
0
Z
Z
Z
π/4
0 √ 2/2
ρ=sec φ ρ4 4 ρ=0
sin φ sec4 φd φ = √
1 4
Z
2π
0
du 1 2 =− π+ π 4 6 3 u
Dr. Lahcen Laayouni
Advanced Calculus
dθ
Z
π/4 0
sin φ dφ cos4 φ
Multiple Integrals
Spherical coordinates
Triple integrals
Example Evaluate I = R
ZZZ
(x 2 + y 2 )dV , where R
is the region inside the sphere
x 2 + y 2 + z 2 = 4a2
and outside the
cylinder x 2 + y 2 = a2 , a > 0 . In spherical coordinates the sphere has the equation x 2 + y 2 + z 2 = ρ2 = 4a2 , or simply ρ = 2a , while the cylinder has the equation x 2 + y 2 = r 2 = (ρ sin φ)2 = a2 , or ρ sin φ = a . Thus the region R is described by 0 ≤ θ ≤ 2π , a sin φ
π 6
≤φ≤
≤ ρ ≤ 2a . Dr. Lahcen Laayouni
Advanced Calculus
π 2
, and
Multiple Integrals
Spherical coordinates
Triple integrals
Example (cont.) In terms of spherical coordinates the required integral is Z 2π Z π/2 Z 2a I=2 dθ sin φd φ ρ2 sin2 φρ2 d ρ . π/6
0
a/ sin φ
Using the cylindrical coordinates the region R is described by √ 0 ≤ θ ≤ 2π , a ≤ r ≤ 2a , and 0 ≤ z ≤ 4a2 − r 2 , and Z Z Z √ 2π
I=2
0
4a2 −r 2
2a
dθ
rdr
r 2 dz ,
0
a
which is much more easier than (1). Thus Z 2π Z 2a p I=2 dθ r 3 4a2 − r 2 dr 0
a
Dr. Lahcen Laayouni
Advanced Calculus
(1)
Multiple Integrals
Spherical coordinates
Triple integrals
Example (cont.) Setting u = 4a2 − r 2 , du = −2rdr , then Z 2π Z 3a2 √ I = dθ (4a2 − u) udu 0 0 ! 3a2 u 3/2 u 5/2 44 √ 2 = 2π 4a − = 3πa5 . 3/2 5/2 5 0
Remark If the domain of integration has spherical and axial symmetry, then we chose cylindrical or spherical coordinates system according to whether the integrand involves x 2 + y 2 or x 2 + y 2 + z 2 term. Dr. Lahcen Laayouni
Advanced Calculus
Advanced Calculus Lecture 4 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 16, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Spherical coordinates In the system of spherical coordinates a point P = [x, y, z] is represented by x y z
= ρ sin φ cos θ, = ρ sin φ sin θ, = ρ cos φ,
0≤φ≤π 0 ≤ θ ≤ 2π . ρ≥0
Note that ρ2 = x 2 + y 2 + z 2 = r 2 + z 2 . The r coordinate in cylindrical coordinates is related to ρ and φ by q r = x 2 + y 2 = ρ sin φ . also
r tan φ = = z
p
x2 + y2 z
Dr. Lahcen Laayouni
and Advanced Calculus
tan θ =
y . x
Multiple Integrals
Spherical coordinates
Triple integrals
The volume element in spherical coordinates The volume element in spherical coordinates is given by dV = ρ2 sin φd ρd φd θ . If f is a continuous function on the bounded region R , then ZZZ ZZZ f (x, y, z)dV = g(ρ, φ, θ)ρ2 sin φd ρd φd θ , R
S
where g(ρ, φ, θ) = f (x, y, z) and S is the region R expressed in spherical coordinates. Remark Spherical coordinates are suited to problems involving spherical symmetry and, in particular, to regions bounded by spheres centered at the origin, circular cones with axes along the z-axis, and vertical planes containing the z-axis. Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example Evaluate
I=
ZZZ
(1 + x 2 + y 2 )dV ,
R
3.0 2.5 2.0
where R is the region in the first octant bounded by two spheres of radius 2 and 3 respectively.
z1.5 1.0 3.0 0.5
2.0
1.5 0.0 0.0 0.0
0.5 0.5
1.0
1.0
2.5
y
1.5 x
2.0
2.5
3.0
In spherical coordinates, the region R is given by the inequalities 2 ≤ ρ ≤ 3 , 0 ≤ φ ≤ π/2 , 0 ≤ θ ≤ π/2 , and the integrand becomes 1 + x 2 + y 2 = 1 + r 2 = 1 + ρ2 sin2 φ . Then Z π/2 Z π/2 Z 3 I = dθ dφ (1 + ρ2 sin2 φ)ρ2 sin φd ρ 0
0
Dr. Lahcen Laayouni
2
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example Thus I =
Z
π/2
dθ
0
= = = =
Z
π/2
dφ
0
Z
3
(ρ2 sin φ + ρ4 sin3 φ)d ρ
2
ρ=3 ρ3 ρ5 3 dθ dφ sin φ + sin φ 3 5 0 0 ρ=2 Z π/2 Z π/2 19 211 dθ sin φ + sin3 φ d φ 3 5 0 0 φ=π/2 Z π/2 517 211 2 dθ − cos φ − sin φ cos φ 15 15 0 φ=0 Z π/2 517 517 dθ = π. 15 0 30 Z
π/2
Z
π/2
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example Evaluate I =
ZZZ q
x 2 + y 2 + z 2 dV ,
E
where E
1.0
is the solid bounded by
0.75 z
z=1,
z 2 = x 2 + y 2 , above the
−1.0
xy -plane.
0.5
0.25 x y −0.5 −0.5 0.0 0.0
−1.0
0.0
0.5
0.5 1.0
1.0
In spherical coordinates the equation of the cone becomes (ρ cos φ)2 = (ρ sin φ cos θ)2 + (ρ sin φ sin θ)2 , which reduces to cos2 φ = sin2 φ(cos2 θ + sin2 θ) = sin2 φ . or cos φ = sin φ , from which we find that tan φ = 1 .Thus the cone is simply defined by 0 ≤ φ ≤ π/4 . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Spherical coordinates
Triple integrals
Example The plane is described by ρ cos φ = 1 , or ρ = sec φ , thus Z 2π Z π/4 Z sec φ I = dθ dφ ρρ2 sin φd ρ 0
= = =
Z Z
0
2π
dθ 0
dθ 0
= −
π/4
Z
π/4
sin φd φ
0
2π
1 4
0
Z
2π
dθ
0
2π 4
Z
1
sec φ
ρ3 d ρ
0
sin φd φ
0
Z
Z
Z
π/4
0 √ 2/2
ρ=sec φ ρ4 4 ρ=0
sin φ sec4 φd φ = √
1 4
Z
2π
0
du 1 2 =− π+ π 4 6 3 u
Dr. Lahcen Laayouni
Advanced Calculus
dθ
Z
π/4 0
sin φ dφ cos4 φ
Multiple Integrals
Spherical coordinates
Triple integrals
Example Evaluate I = R
ZZZ
(x 2 + y 2 )dV , where R
is the region inside the sphere
x 2 + y 2 + z 2 = 4a2
and outside the
cylinder x 2 + y 2 = a2 , a > 0 . In spherical coordinates the sphere has the equation x 2 + y 2 + z 2 = ρ2 = 4a2 , or simply ρ = 2a , while the cylinder has the equation x 2 + y 2 = r 2 = (ρ sin φ)2 = a2 , or ρ sin φ = a . Thus the region R is described by 0 ≤ θ ≤ 2π , a sin φ
π 6
≤φ≤
≤ ρ ≤ 2a . Dr. Lahcen Laayouni
Advanced Calculus
π 2
, and
Multiple Integrals
Spherical coordinates
Triple integrals
Example (cont.) In terms of spherical coordinates the required integral is Z 2π Z π/2 Z 2a I=2 dθ sin φd φ ρ2 sin2 φρ2 d ρ . π/6
0
a/ sin φ
Using the cylindrical coordinates the region R is described by √ 0 ≤ θ ≤ 2π , a ≤ r ≤ 2a , and 0 ≤ z ≤ 4a2 − r 2 , and Z Z Z √ 2π
I=2
0
4a2 −r 2
2a
dθ
rdr
r 2 dz ,
0
a
which is much more easier than (1). Thus Z 2π Z 2a p I=2 dθ r 3 4a2 − r 2 dr 0
a
Dr. Lahcen Laayouni
Advanced Calculus
(1)
Multiple Integrals
Spherical coordinates
Triple integrals
Example (cont.) Setting u = 4a2 − r 2 , du = −2rdr , then Z 2π Z 3a2 √ I = dθ (4a2 − u) udu 0 0 ! 3a2 u 3/2 u 5/2 44 √ 2 = 2π 4a − = 3πa5 . 3/2 5/2 5 0
Remark If the domain of integration has spherical and axial symmetry, then we chose cylindrical or spherical coordinates system according to whether the integrand involves x 2 + y 2 or x 2 + y 2 + z 2 term. Dr. Lahcen Laayouni
Advanced Calculus
