Rectangular vs. Cylindrical vs. Spherical –MATH 2421
Spring 2006
Mike Kawai
Consider the most popular (in terms of quizzes and tests) surfaces and solids
using Rectangular, Cylindrical, and Spherical Coordinates...
(#1) We always give you the basic transformations.
(a) Rectangular ! Polar
r=
p
x2 + y 2
tan ( ) =
It is highly desirable to have r
y
x
[You choose the correct Quadrant for !]
0:
(b) Polar ! Rectangular
x = r cos ( )
y = r sin ( )
(c) Cylindrical Coordinates merely copy the z-coordinate in 3-D.
(x; y; z) $ (r; ; z) :
Change Rectangular to Polar, or vice versa.
(d) Rectangular ! Spherical
This is the 3-D analog to 2-D Polar.
We have the distance from the origin, ; and two angles which determine direction,
:
p
= x2 + y 2 + z 2
and
When we locate a point in 3-D Spherical ( ; ; ) ; we …rst …nd the half-plane associated
with :
For example, when = 0; we have the xz-plane, for x 0: [See below, left.] We show
that all of these planes are perpendicular to the xy-plane!
Above, right, we also see the planes
=
1
6
;
4
;
3
;
2
:
(i) Since these planes are perpendicular to the xy-plane, their equations in Rectangular
Coordinates must only contain x and y:
For example, the plane
=
6
is equivalent to
tan ( ) = tan
6
p
y
3
) =
)y=
x
3
p !
3
x:
3
We know that this process won’t work for = since tan
2
2
know from geometry this is the yz-plane, and its equation is
is unde…ned, but we
x = 0:
So once we have located the the appropriate -plane, we can determine :
(ii) When = 0; we are pointing straight up the positive z-axis. Within the -plane,
we
proceed down toward the xy-plane where
=
2
:
Consider the following set-up:
Some texts refer to my -plane as the “rz-plane”for a given value of : For example,
if we have the Rectangular Coordinates
p
(x; y; z) = 1; 0; 3 ;
y
= 0 ) = 0:
x
Within the half-plane = 0; we can calculate the Cylindrical Coordinates
we see that tan ( ) =
p
(r; ; z) = 1; 0; 3
p
p
because r = x2 + y 2 = 12 + 02 = 1: I will sketch in a line segment in that plane.
[In this case, the x-axis is the same as the r-axis! We could have done something
similar with the y-axis.]
z
2
The line segment
pextends from the origin to
(x; y; z) = 1; 0; 3 :
1
0
-2
-1
0
1
2
Label the angle from the z-axis down to the line segment
(clockwise) with the letter :
r
-1
The p
length of the line segment is
= x2 + y 2 + z 2 = 2:
-2
2
Now if we project the line segment onto the z-axis,
we see that the z-coordinate must be:
z = cos ( ) :
2
z
1
Similarly, we have
r = sin ( ) :
0
-2
-1
0
1
2
r
-1
Since
x = r cos ( ) and y = r sin ( ) ; we must have
x = sin ( ) cos ( )
y = sin ( ) sin ( ) :
-2
That completes our transformation from Spherical to Rectangular.
(e) Spherical ! Rectangular
We use the same relationships from above.
p
=
x2 + y 2 + z 2
tan ( ) =
cos ( ) =
y
x
z
[same as Polar]
z
=p
x2
+
y2
+
z2
)
= cos
1
z
We note that the inverse cosine function can only yield 0
; which is the correct
maximal domain for :
Again, we remind you that is never negative and we must have 0
<2 :
(#2) The great majority of the “interesting” surfaces will be symmetric about the z-axis. That
is, they will be surfaces of revolution. This …rst group of surfaces does NOT need to be
symmetric about the z-axis.
Cylindrical surfaces.
Any surface whose equation in 3-D Rectangular Coordinates has two or fewer variables must
be a cylindrical surface.
p
3 x :
We’ve already seen planes which are perpendicular to the xy-plane y = x; y =
If an equation has precisely two variables represented, then in the plane de…ned by those two
variables (in this case, the xy-plane), we have the generating curve. In the case of the 3-D
equation y = x; we have the line y = x in the xy-plane. Since z is not mentioned, we allow
z to take on any real value. Thus, we duplicate the generating curve in the z-axis direction
and obtain the plane y = x:
(a) Suppose we have a circle in the xy-plane,
x2 + y 2 = 9:
If we extend this to 3-D, then we create a right circular cylinder. The circle is duplicated
in the z-axis direction.
3
Where does the “right” come from? If the cylindrical surface is perpendicular to the
plane which contains the generating curve, then right angles are clearly formed. If
the surface is not at right angles to that plane, then we say it is an oblique cylindrical
surface.
[See next page. The …gure on the left is the right circular cylinder, and the …gure on
the right is the oblique circular cylinder.]
Here’s an example of a right circular cylinder which is perpendicular to the yz-plane.
y2 + z2 = 9
Recall that if a surface is a function surface, z = f (x; y) ; it must pass the equivalent of
the Vertical Line Test in 3-D.
Clearly, the surface on the left does not pass that test.
However, we solve for z and take the upper half only, we have
p
f (x; y) = 9 y 2 ;
the upper half of the right circular cylinder perpendicular to the yz-plane.
4
(b) Below, left, we have a right parabolic cylinder perpendicular to the xy-plane.
x=
y2:
Below, right, we have a right hyperbolic cylinder perpendicular to the xz-plane.
x2
z 2 = 1:
Note that we also often plot the xy-plane as a reference (try to ignore it).
(#3) More Planes...
(a) In Spherical Coordinates, the xy-plane, z = 0; is actually a ‡attened cone [later]. We
have
= ) cos ( ) = cos
= 0:
2
2
Now multiply both sides by :
cos ( ) =
0 = 0 ) z = 0:
(b) If the plane is parallel to the xy-plane, but does not pass through the origin, then we
have something like z = 1:
We can transform this into Spherical Coordinates.
z = 1 ) cos ( ) = 1 )
=
1
:
cos ( )
In the rz-plane, the generating curve is the line z = 1: The limits for
=
1
; 0
cos ( )
<
2
; 0
Whenever we use Spherical Coordinates, we want
are tricky...
2 :
as a function of
and :
(c) Similarly, if we have the plane x = 1; we have
x = 1 ) sin ( ) cos ( ) = 1 )
=
5
1
; 0<
sin ( ) cos ( )
< ;
2
<
<
2
: