Section 12.6
Cylinders and Quadric Surfaces
A cylinder is a three dimensional shape that is determined by
• a two dimensional (plane) curve C in three dimensional space
• a line L in a plane not parallel to the one in which the C lies.
The cylinder is the set of all lines passing through C that are parallel to L.
For example, consider the curve and line in the graph below:
These two graphs determine the cylinder in blue:
This surface is actually the graph of the z = y 2 in three dimensional space. Note that the
equation z = y 2 puts no restrictions on the variable x. In other words, the surface is the set of
all points (x, y, y 2 ); so if we determine the two dimensional shape of the set of points of the form
(y, y 2 ) (obviously a parabola), then we can move this shape along the x axis to generate the surface
z = y 2 . In particular, if we make a specific choice for x, say x = 2, we look at the set of points of
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Section 12.6
the form (2, y, y 2 ) (plotted in black), which has exactly the same shape as if we were to look at the
graph at x value −1, where we would see the points (−1, y, y 2 ) (plotted in yellow):
Example
Graph the three dimensional surface defined by y 2 + z = 4.
Note that the equation y 2 + z = 4 puts no restrictions on the variable x. Again, this means
that determining the shape of the curve y 2 + z = 4 in the yz plane will help us find the shape of
the three dimensional surface. We can rewrite y 2 + z = 4 as z = 4 − y 2 , which is easy to graph in
two dimensions (again, in the yz plane):
Now the graph of the three dimensional surface generated by y 2 + z = 4 is the set of all points
(x, y, 4 − y 2 ); again, x is completely independent of y, so we expect to see the same shape y 2 + z = 4
at any point x. To create the three dimensional graph, we can think of sliding the two dimensional
graph of y 2 + z = 4 along the x axis:
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Section 12.6
We will occasionally need to know equations for quadric surfaces. A quadric surface is a threedimensional shape described by a second-degree equation in x, y, and z. In this section, we will
discuss the equations related to each of these three-dimensional shapes.
Ellipsoid
An ellipsoid is described by an equation of the form
x2 y 2 z 2
+ 2 + 2 = 1,
a2
b
c
where a, b, and c are constants.
Elliptic Parabaloid
An elliptic parabaloid is described by an equation of the form
z
x2 y 2
+ 2 = ,
a2
b
c
where a, b, and c are constants.
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Section 12.6
Elliptic Cone
An elliptic cone is described by an equation of the form
x2 y 2
z2
+
=
,
a2
b2
c2
where a, b, and c are constants.
Hyperboloid of One Sheet
A hyperboloid of one sheet is described by an equation of the form
x2 y 2 z 2
+ 2 − 2 = 1,
a2
b
c
where a, b, and c are constants.
Hyperboloid of Two Sheets
A hyperboloid of two sheets is described by an equation of the form
z 2 x2 y 2
− 2 − 2 = 1,
c2
a
b
where a, b, and c are constants.
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Section 12.6
Hyperbolic Parabaloid
A hyperbolic parabaloid is described by an equation of the form
z
y 2 x2
− 2 = ,
2
b
a
c
where a, b, and c are constants and c > 0.
Graphing a quadric surface can be quite challenging, but we can make the process a bit simpler
by first identifying the type of surface given by the equation. In order to generate a graph, it is
helpful to graph traces, i.e. two dimensional slices of the surface.
Example
Sketch a graph of the surface z 2 = x2 +
y2
4 .
Comparing the equation with those of the quadric surfaces that we have seen, it appears that
the surface should be an elliptic cone. Let’s try drawing some of the traces of the curve, first for a
few different values of z, say z = 0, z = 2, and z = −2.
At height z = 0, the two dimensional shape is given by the set of all points (x, y) satisfying
2
2
x + y4 = 0. The only point satisfying this equation is (0, 0), so the trace at z = 0 is just the point
(0, 0, 0).
At height z = 2, the two dimensional shape is given by the set of all points (x, y) satisfying
2
2
x + y4 = 4. This is just an ellipse passing through the points (±2, 0) and (0, ±4):
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Section 12.6
x2
The trace at z = −2 is the two dimensional shape given by the set of all points (x, y) satisfying
2
+ y4 = 4, which is the same ellipse as in the previous example.
Graphing these traces in three dimensions gives us the following picture:
From this, it is fairly simple to sketch the curve itself:
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Section 12.6
Challenge problem
Let L1 and L2 be a pair of perpendicular skew lines (i.e. lines that are perpendicular but
non-intersecting, as they lie in parallel planes).
One of the quadric surfaces is precisely the set of all points that are equidistant from the pair
of lines. Determine which quadric surface this is, and show that your answer is correct.
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