In this section we will introduce spherical
coordinates. Spherical coordinates can
take a little getting used to. It’s
probably easiest to start things off with a sketch.
Spherical coordinates consist of the following three
quantities.
First there is . This is the distance from the origin to the
point and we will require .
Next there is . This is the same angle that we saw in
polar/cylindrical coordinates. It is the
angle between the positive xaxis and
the line above denoted by r (which is
also the same r as in
polar/cylindrical coordinates). There
are no restrictions on .
Finally there is . This is the angle between the positive zaxis and the line from the origin to
the point. We will require .
In summary, is the distance from the origin to the point, is the angle that we need to rotate down from
the positive zaxis to get to the
point and is how much we need to rotate around the zaxis to get to the point.
We should first derive some conversion formulas. Let’s first start with a point in spherical
coordinates and ask what the cylindrical coordinates of the point are. So, we know and want to find . Of course we really only need to find r and z since is the same in both coordinate systems.
We will be able to do all of our work by looking at the
right triangle shown above in our sketch.
With a little geometry we see that the angle between z and is and so we can see that,
and these are exactly the formulas that we were looking
for. So, given a point in spherical
coordinates the cylindrical coordinates of the point will be,
Note as well that,
Or,
Next, let’s find the Cartesian coordinates of the same
point. To do this we’ll start with the
cylindrical conversion formulas from the previous
section.
Now all that we need to do is use the formulas from above
for r and z to get,
Also note that since we know that we get,
Converting points from Cartesian or cylindrical coordinates
into spherical coordinates is usually done with the same conversion
formulas. To see how this is done let’s
work an example of each.
Example 1 Perform
each of the following conversions.
(a) Convert
the point from cylindrical to spherical coordinates.
[Solution]
(b) Convert
the point from Cartesian to spherical coordinates.
[Solution]
Solution
(a) Convert the point from cylindrical to spherical coordinates.
We’ll start by acknowledging that is the same in both coordinate systems and
so we don’t need to do anything with that.
Next, let’s find .
Finally, let’s get . To do this we can use either the conversion
for r or z. We’ll use the
conversion for z.
Notice that there are many possible values of that will give ,
however, we have restricted to the range and so this is the only possible value in
that range.
So, the spherical coordinates of this point will are .
[Return to Problems]
(b) Convert the point from Cartesian to spherical coordinates.
The first thing that we’ll do here is find .
Now we’ll need to find . We can do this using the conversion for z.
As with the last parts this will be the only possible in the range allowed.
Finally, let’s find . To do this we can use the conversion for x or y. We will use the
conversion for y in this case.
Now, we actually have more possible choices for but all of them will reduce down to one of
the two angles above since they will just be one of these two angles with one
or more complete rotations around the unit circle added on.
We will however, need to decide which one is the correct
angle since only one will be. To do
this let’s notice that, in two dimensions, the point with coordinates and lies in the second quadrant. This means that must be angle that will put the point into
the second quadrant. Therefore, the
second angle, ,
must be the correct one.
The spherical coordinates of this point are then .
[Return to Problems]

Now, let’s take a look at some equations and identify the
surfaces that they represent.
Example 2 Identify
the surface for each of the following equations.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
There are a couple of ways to think about this one.
First, think about what this equation is saying. This equation says that, no matter what and are, the distance from the origin must be
5. So, we can rotate as much as we
want away from the zaxis and
around the zaxis, but we must
always remain at a fixed distance from the origin. This is exactly what a sphere is. So, this is a sphere of radius 5 centered
at the origin.
The other way to think about it is to just convert to
Cartesian coordinates.
Sure enough a sphere of radius 5 centered at the origin.
[Return to Problems]
(b)
In this case there isn’t an easy way to convert to
Cartesian coordinates so we’ll just need to think about this one a
little. This equation says that no
matter how far away from the origin that we move and no matter how much we
rotate around the zaxis the point
must always be at an angle of from the zaxis.
This is exactly what happens in a cone. All of the points on a cone are a fixed
angle from the zaxis. So, we have a cone whose points are all at
an angle of from the zaxis.
[Return to Problems]
(c)
As with the last part we won’t be able to easily convert
to Cartesian coordinates here. In this
case no matter how far from the origin we get or how much we rotate down from
the positive zaxis the points must
always form an angle of with the xaxis.
Points in a vertical plane will do this. So, we have a vertical plane that forms an
angle of with the positive xaxis.
[Return to Problems]
(d)
In this case we can
convert to Cartesian coordinates so let’s do that. There are actually two ways to do this
conversion. We will look at both since
both will be used on occasion.
Solution 1
In this solution method we will convert directly to
Cartesian coordinates. To do this we
will first need to square both sides of the equation.
Now, for no apparent reason add to both sides.
Now we can convert to Cartesian coordinates.
So, we have a cylinder of radius 2 centered on the zaxis.
This solution method wasn’t too bad, but it did require
some not so obvious steps to complete.
Solution 2
This method is much shorter, but also involves something
that you may not see the first time around.
In this case instead of going straight to Cartesian coordinates we’ll
first convert to cylindrical coordinates.
This won’t always work, but in this case all we need to do
is recognize that and we will get something we can
recognize. Using this we get,
At this point we know this is a cylinder (remember that
we’re in three dimensions and so this isn’t a circle!). However, let’s go ahead and finish the
conversion process out.
[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.
This won’t always be easier, but it can make some of the conversions
quicker and easier.
The last thing that we want to do in this section is
generalize the first three parts of the previous example.