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Spherical
Coordinates

Spherical Coordinates
In this section we will introduce spherical coordinates
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It’s probably easiest to start things off with a sketch
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First there is ρ
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Next there is θ
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It is the
angle between the positive x-axis and the line above denoted by r (which is also the same r as in
polar/cylindrical coordinates)
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Finally there is ϕ
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We will require 0 ≤ ϕ ≤ π
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We should first derive some conversion formulas
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So, we know ( ρ , θ , ϕ ) and
what to find ( r, θ , z )
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We will be able to do all of our work by looking at the right triangle shown above in our sketch
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So, given a point in spherical
coordinates the cylindrical coordinates of the point will be,

r = ρ sin ϕ

θ =θ
z = ρ cos ϕ
Note as well that,

r 2 + z 2 = ρ2 cos 2 ϕ + ρ2 sin 2 ϕ = ρ2 ( cos 2 ϕ + sin 2 ϕ ) = ρ2

Or,

ρ2 = r 2 + z 2
Next, let’s find the Cartesian coordinates of the same point
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x = r cos θ
y = r sin θ
z=z
Now all that we need to do is use the formulas from above for r and z to get,

x = ρ sin ϕ cos θ
y = ρ sin ϕ sin θ
z = ρ cos ϕ
Also note that since we know that r 2 = x 2 + y 2 we get,

ρ2 = x 2 + y 2 + z 2
Converting points from Cartesian or cylindrical coordinates into spherical coordinates is usually
done with the same conversion formulas
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Example 1 Perform each of the following conversions
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4
⎝
⎠
[Solution]

(

)

(b) Convert the point −1,1, − 2 from Cartesian to spherical coordinates
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4
⎝
⎠
We’ll start by acknowledging that θ is the same in both coordinate systems and so we don’t need
to do anything with that
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ρ = r + z = 6+2 = 8 = 2 2
2

2

Finally, let’s get ϕ
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We’ll use the
conversion for z
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z

2
cos ϕ = =
ρ 2 2

−1

π π⎞
⎛
So, the spherical coordinates of this point will are ⎜ 2 2, , ⎟
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The first thing that we’ll do here is find ρ
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We can do this using the conversion for z
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Finally, let’s find θ
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We will use the conversion
for y in this case
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We will however, need to decide which one is the correct angle since only one will be
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This means that θ must be angle that will put the point into the second
quadrant
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⎛ 3π 3π ⎞
, ⎟
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Example 2 Identify the surface for each of the following equations
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First, think about what this equation is saying
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So, we can rotate as much as we want away from the
z-axis and around the z-axis, but we must always remain at a fixed distance from the origin
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So, this is a sphere of radius 5 centered at the origin
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ρ =5

ρ 2 = 25
x 2 + y 2 + z 2 = 25
Sure enough a sphere of radius 5 centered at the origin
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This equation says that no matter how far away from the origin that we
move and no matter how much we rotate around the z-axis the point must always be at an angle
of π3 from the z-axis
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All of the points on a cone are a fixed angle from the z-

axis
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[Return to Problems]

2π
(c) θ =
3
As with the last part we won’t be able to easily convert to Cartesian coordinates here
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Points in a vertical plane will do this
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[Return to Problems]

(d) ρ sin ϕ = 2
In this case we can convert to Cartesian coordinates so let’s do that
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We will look at both since both will be used on occasion
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To do this we will first
need to square both sides of the equation
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ρ 2 sin 2 ϕ + ρ 2 cos 2 ϕ = 4 + ρ 2 cos 2 ϕ
ρ 2 ( sin 2 ϕ + cos 2 ϕ ) = 4 + ρ 2 cos 2 ϕ
ρ = 4 + ( ρ cos ϕ )
2

2

Now we can convert to Cartesian coordinates
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This solution method wasn’t too bad, but it did require some not so obvious steps to complete
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In this case instead of going straight to Cartesian coordinates we’ll first convert to
cylindrical coordinates
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Using this we get,

ρ sin ϕ = 2
r

2

At this point we know this is a cylinder (remember that we’re in three dimensions and so this isn’t
a circle!)
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r2 = 4
x +y =4
2

2

[Return to Problems]

So, as we saw in the last part of the previous example it will sometimes be easier to convert
equations in spherical coordinates into cylindrical coordinates before converting into Cartesian
coordinates
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The last thing that we want to do in this section is generalize the first three parts of the previous
example

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