Section 15.9
Triple Integrals in Spherical Coordinates
Another way to represent points in 3 dimensional space is via spherical coordinates, which write
⃗ , i.e. the distance from the
a point P as P = (ρ, θ, ϕ). The number ρ is the length of the vector OP
origin to P :
In particular, since ρ is a distance, it is never negative.
The remaining coordinates θ and ϕ are both angles; θ is just the angle that we have used in
⃗ makes with
cylindrical coordinates. However, ϕ is new to us: it is the angle that the vector OP
the positive part of the z axis:
For example, the point (2, π3 , π6 ) is graphed below:
Using spherical coordinates instead of rectangular coordinates can greatly simplify the equations
of many types of shapes. For example, the graph of the equation ϕ = π3 is a cone:
1
Section 15.9
The graph of ρ = 2 is a sphere of radius 2, and the graph of θ = π4 is a plane through the z axis.
Spherical and conical shapes are generally best represented using spherical coordinates.
We will often need to convert between Cartesian and spherical coordinates or between cylindrical
and spherical coordinates; here are the conversions:
x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ,
√
√
r = ρ sin ϕ, ρ = x2 + y 2 + z 2 = r2 + z 2 .
Integrating in Spherical Coordinates
In order to integrate the function f (x, y, z) over the region E in spherical coordinates, we will
need to begin by rewriting f (x, y, z) as a function of ρ, θ, and ϕ using the conversions above.
Next, sketch a graph of the region E, and rewrite the bounds on E in spherical coordinates.
Find the bounds on ρ by passing line segments from the origin through the region; the surface
g1 (θ, ϕ) where the segments enter E is the lower bound on ρ, and the surface g2 (θ, ϕ) is the upper
bound on ρ.
Next we need to determine the bounds on ϕ; think of dropping line segments down through
the origin from the positive part of the z axis through the region. The surface h1 (θ) where the
segments enter E is the lower bound on ϕ, and the surface h2 (θ) where the segments leave E is the
upper bound on ϕ.
Finally, look at the ”shadow” of E in the xy plane, and determine the bounds on θ as we would
in cylindrical coordinates.
Theorem 0.0.1. The triple integral of the function f (x, y, z) over the region E described in
spherical coordinates by
g1 (θ, ϕ) ≤ ρ ≤ g2 (θ, ϕ), h1 (θ) ≤ ϕ ≤ h2 (θ), and α ≤ θ ≤ β
is
∫
β
α
∫
h2 (θ) ∫ g2 (θ,ϕ)
h1 (θ)
f (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ)ρ2 sin ϕ dρdϕdθ.
g1 (θ,ϕ)
2
Section 15.9
Notice that, as with cylindrical coordinates, we must multiply the function f by an extra
factor (in this case, ρ2 sin ϕ) in order to account for the fact that we are integrating in spherical
coordinates.
Examples
Find the volume of the solid that lies inside the sphere x2 + y 2 + z 2 = 2 and outside the cone
z 2 = x2 + y 2 .
Since we want to use triple integrals to find the volume of a solid in three dimensional space,
we want to integrate the function 1 over the region described above. However, it will be much
simpler in this case to integrate in spherical coordinates, so when we convert our functions we must
multiply the integrand by ρ2 sin ϕ, as indicated in the theorem; so to find the volume of the region,
we will actually integrate the function ρ2 sin ϕ.
To find the bounds on the integrals, we need to√start by sketching the region described above.
The surface x2 + y 2 + z 2 = 2 is a sphere of radius 2 centered at the origin:
To graph the surface z 2 = x2 + y 2 , we use level curves. The level curve at height z = 0 is the
set of point (x, y, 0) so that x2 + y 2 = 0. In other words, the level curve at height 0 is just the
origin. The level curve at height z = 2 is the set of points (x, y, 2) so that x2 + y 2 = 4, i.e. the
circle of radius 2 at height z = 2. Similarly, the level curve at height z = −2 is the circle of radius
2 at height z = −2. The cone is graphed below:
3
Section 15.9
Putting the two graphs together, we want to find the volume of the region sketched below:
Let’s find the bounds
√ on ρ. Note that line segments from the origin through the region always
vary from ρ = 0 to ρ = 2, so these numbers form the bounds on ρ.
We next need to find the bounds on ϕ. Notice that, at each point on the top surface of the
region, the value for ϕ is the same (and a similar statement is true for the bottom surface of the
region); so to find ϕ, we only have to look at a specific case.
4
Section 15.9
z2
Let’s think about the graph of z 2 = x2 + √
y 2 in the yz plane, i.e. when x = 0. Then we have
2
= y , and since the radius of the sphere is 2, we end up with the following triangle:
It should be clear that the angle above is π4 . Thus the lower bound on ϕ is also
similar argument we see that the upper bound on ϕ is 3π
4 .
Finally, let’s look at the projection of the region into the xy plane.
The values for θ extend from 0 to 2π.
5
π
4,
and using a
Section 15.9
So the volume of the region is given by
∫
2π
∫
3π/4 ∫
√
2
ρ2 sin ϕ dρdϕdθ.
0
π/4
0
Let’s evaluate the integral:
∫
2π
∫
3π/4 ∫
√
2
)
√2
1
( ρ3 sin ϕ)0
dϕdθ
3
0
π/4
∫ 2π ∫ 3π/4 √
2 2
=
sin ϕ dϕdθ
3
0
π/4
√
)
∫ 2π (
3π/4
2 2
(−
=
cos ϕ) π/4
dθ
3
0
∫ 2π √ √
2 2
=
( 2) dθ
3
0
∫ 2π
4
=
dθ
3
0
4 2π
= θ 0
3
8π
.
=
3
∫
ρ2 sin ϕ dρdϕdθ =
0
π/4
0
6
2π
∫
3π/4 (