Sect. 13.6 (Cylinders & Quadric Surfaces)

Sect. 13.6 (Cylinders & Quadric Surfaces) —MATH 2421

Kawai

Sect. 13.6 (Cylinders & Quadric Surfaces) 1. Cylindrical surfaces. If we have the circle in the xy-plane: x2 + y2 = 4, then what happens if we use this equation in 3D? Answer: We say that z can do anything. Thus, we obtain a surface, and the only constraint is that the projection down onto the xy-plane must be that circle. We get the inﬁnte tennis ball can. We say that the generating curve is in the xy-plane. Thus, the equation will only have x and y in it, and z is allowed to take on any real value.

We say that this is a “right”circular cylinder because the generating curve is copied at right angles to the plane that contains the generating curve. This copying process is called extrusion, and we say that the right cylinder is extruded from the xy-plane. We create rigatoni pasta this way.

In this course, we only care about cylindrical surfaces that are orthogonal to simple planes. Thus, we have two more possible orientations:

x2 + z2 = 4 AND y2 + z2 = 4.

[You try it!]

2. Cylindrical surfaces can come in a variety of shapes.

(a) Parabolic cylinders: y = x2, x = y2, etc. There are 6 possible orientations. (b) Sinusoidal cylinders: y = sin (x) , etc. (c) Planes can be regarded as cylindrical surfaces (walls). The line y = x is in the xy-plane, but the plane y = x is a vertical plane, orthogonal to the xy-plane. In theory, this could be classiﬁed as a right cylindrical wall.

3. I have attached the handout on Conic Sections and my signature lecture on Quadric Surfaces. Enjoy!

1 Fun(?) with Quadric Surfaces MATH 2421 –Quadrics.tex

Mike Kawai

The General Quadric Surface... is a huge mess.

Ax2 + By2 + Cz2 + Dxy + Exz + F yz + Gx + Hy + Iz + J = 0:

We’llignore surfaces which have xy; xz; or yz terms in them. Exception: You should know that z = xy is a hyperbolic paraboloid (saddle shape), as you will see later.

(#1) Ellipsoid. The standard form for this course is

x2 y2 z2 + + = 1: a2 b2 c2

We assume that the center of the ellipse is the origin. If the center is located at (x0; y0; z0) ; then we must translate the ellipse accordingly:

2 2 2 (x x0) (y y0) (z z0) + + = 1: a2 b2 c2 In terms of coursework, the center is almost always at the origin, but be prepared to complete the square, just in case!

(a) Why is this surface called an “ellipsoid”? Our nomenclature system is based on a democratic voting system. We must examine the traces parallel to the coordinate planes. (Recall that a “trace”is the curve which results from intersecting a plane with a surface.) Example: x2 y2 z2 + + = 1: 4 9 16

This is the standard view for this course. The positive x-axis appears to come out of the paper.

Basically, it looks like a stretched out sphere.

Let’slook at the traces by projecting the surface down onto the coordinate planes.

1 Here is the view if we smash the surface down onto the yz-plane (x = 0) :

By substituting x = 0 into the equation, we obtain

y2 z2 + = 1: 9 16

This is a “3 by 4”ellipse in the yz-plane! The semiminor axis (half of the shorter axis) has length 3 and is parallel to the y-axis. The semimajor axis has length 4 and is parallel to the z-axis.

Smash it down onto the xz-plane (y = 0) :

x2 z2 + = 1: 4 16

This is a “2 by 4”ellipse in the xz-plane! The semiminor axis (half of the shorter axis) has length 2 and is parallel to the x-axis. The semimajor axis has length 4 and is parallel to the z-axis.

Finally, smash it down onto the xy-plane (z = 0) :

x2 y2 + = 1: 4 9

This is a “2 by 3”ellipse in the xy-plane! The semiminor axis (half of the shorter axis) has length 2 and is parallel to the x-axis. The semimajor axis has length 3 and is parallel to the y-axis.

Since all three traces are ellipses, we have three votes for “ellipse”, and the surface is given the name “ellipsoid”. In mathematics, the su¢ x “oid”usually refers to a surface, but there are exceptions (a “cardioid”is polar curve).

2 (b) Many of the surfaces we encounter will be surfaces of revolution about the z-axis. Typically, we take a generating curve in the xz-plane or the yz-plane, and then revolve it about the z-axis to form the surface of revolution. x2 z2 For example, in the xz-plane, suppose we have the “2 by 4”ellipse, + = 1: 4 16 The right half of the ellipse is

x2 z2 z2 + = 1 x = 2 1 : 4 16 ) r 16

y 4

If we revolve this about the z-axis, we will obtain

2 a “2 by 4”ellipse trace in the xz-plane AND in the yz-plane.

0 Thus, the equation of the surface must be -4 -2 0 2 4 x x2 y2 z2 + + = 1: 4 4 16 -2 This type of ellipsoid will have matching numbers below the x2 and y2 terms.

-4

The traces parallel to the xy-plane are circles!

As we will see later, any plane which contains the z-axis will give us an identical trace, the “2 by 4”ellipse!

(c) This surface will NOT pass the 3-D version of the “Vertical Line Test”. If we take a 3-D line parallel to the z-axis, then it will most likely intersect the ellipsoid at two points. One point will be on the “upper” half and the other will be on the “lower”half. Thus, we see that we can divide this surface into two halves:

x2 y2 x2 y2 z = 4 1 and z = 4 1 : r 4 4 r 4 4

3 Often, we simplify some of the fractions... x2 y2 z2 + + = 1 4x2 + 4y2 + z2 = 16 4 4 16 )

z2 = 16 4x2 4y2 = 4 4 x2 y2 z = 2 4 x2 y2: p

We write this as a two-variable function.

z = f (x; y) = 2 4 x2 y2: p The domain of this function must be a subset of R2: We want the contents of the square root to be nonnegative. Thus, we have 4 x2 y2 0 or x2 + y2 4: This is the region inside and including the circle of radius 2, centered at the origin.

Clearly, from the 3-D graph, we see that the range must be 0 z 4 (or [0; 4] in interval notation). x2 y2 z2 (d) We say that + + = 1 is an implicit equation because theoretically, we could 4 4 16 break the surface into function surfaces. In this case, we know that we can divide this surface into the upper and lower ellipsoids. Both of those surfaces will be proper 2-variable functions of x and y: z = f (x; y) = 2 4 x2 y2 [upper ellipsoid] z = f (x; y) = p2 4 x2 y2 [lower ellipsoid] These are the explicit functions becausep we can speci…cally write out the formula for z: Crazier surfaces like z4 + sin (z) + x + y = 1 CANNOT be written as z = something:

Hence, it must always be written in its implicit form and we must rely on technology to plot the 3-D surface.

It isn’tpretty, but it is an accurate characterization of the equation!

4 (#2) Elliptic Paraboloid. The standard form for this course is

x2 y2 z = + : a2 b2 We assume that a; b > 0:

(a) Why is this surface called an “elliptic paraboloid”? x2 y2 Here’sthe standard view for z = + : 1 4

We describe this as “bowl-shape upward”because the concave portion opens upward.

Consider the y = 0 trace.

We obtain the parabola z = x2:

5 Consider the x = 0 trace.

y2 We obtain the parabola z = : 4 1 Note: Multiplying by shrinks the vertical axis and 4

makes the parabola look fatter.

We notice that since the surface is bowl-shape upward, both parabolas must also be concave upward.

Examining the xy-plane (z = 0) wouldn’treally help us since the solution to x2 y2 0 = + is x = 0 and y = 0: 1 4 So if we try z = 1; we obtain an ellipse.

x2 y2 + = 0 gives us a “1 by 2”ellipse. 1 4

So that’stwo votes for parabola and one vote for ellipse (or possibly a circle)...

This surface is called an elliptic paraboloid. Majority rules!

(b) The vertex of the surface (in this case, the lowest point) is located at (0; 0; 0) : It corresponds to the vertices of the two parabolic traces. This is clearly a function surface: x2 y2 z = f (x; y) = + 1 4 Its domain is the entire real plane, R2 and it range is [0; + ) because it has its absolute minimum point at (0; 0; 0) : 1 (c) Most likely, you will play with elliptic paraboloids which are surfaces of revolution about the z-axis. Thus, the traces parallel to the xy-plane will be circles: z = k x2 + y2 : When k = 1; we have the “standard”elliptic (circular) paraboloid.

6 So let’sconsider the orientation of our surface... (i) z = x2 + y2: Bowl-shape upward.

(ii) z = x2 + y2 : Bowl-shape downward. (iii) y = x2 + z2: Bowl-shape to the right. (See below, left.)

(iv) y = x2 + z2 : Bowl-shape to the left. (v) x = y2 + z2: Bowl-shape forward. (See above, right.)

(vi) x = y2 + z2 : Bowl-shape backward. (d) Don’tforget that the basic transforms work on surfaces, also. If we take the basic elliptic paraboloid, z = x2 + y2; re‡ect it across the xy-plane, and then shift it up 3 units, we would have

z = x2 + y2 + 3 z = x2 y2 + 3:

I sketched in the plane z = 0; also.

We can see that the surface protrudes above the xy-plane. The absolute maximum point is located at (0; 0; 3) :

Thus, the gumdrop shape can be described by the inequalities:

0 z x2 y2 + 3:

7 (e) We can translate the vertex of the elliptic paraboloid, but we probably would not need to actually make a 3-D graph. For example, the translated form

z = f (x; y) = (x 3)2 + (y + 2)2 has it absolute minimum point when x = 3 and y = 2: Thus, it is located at (3; 2; 0) : (#3) Hyperbolic Paraboloid. This is often called the “saddle-shape”. The standard form for this course is x2 y2 y2 x2 z = or z = a2 b2 b2 a2 We assume that a; b > 0:

(a) Why is this surface called a “hyperbolic paraboloid”? Here’sthe standard view for z = y2 x2:

Consider the trace x = 0: We have the concave up parabola z = y2:

Consider the trace y = 0: We have the concave down parabola z = x2: I have sketched these below in 2-D.

z 4 z 4

2 2

0 0 -4 -2 0 2 4 -4 -2 0 2 4

y x

-2 -2

-4 -4

8 Now slice the surface with the plane z = 0: This gives us 0 = y2 x2 0 = (y x)(y + x) y = x or y = x: ) ) This is two intersecting lines! It turns out that this …gure is really a degenerate hyperbola. If we consider the trace z = 1; we have 1 = y2 x2; a hyperbola, and the trace z = 1 gives us 1 = y2 x2 1 = x2 y2; which is the other hyperbola. )

y 4 y 4

2 2

0 0 -4 -2 0 2 4 -4 -2 0 2 4

x x

-2 -2

-4 -4

As an experiment, you might try letting y2 x2 = k as k 0+: ! When k is really small, the two branches converge on the origin. For example, look at

y2 x2 = 0:01:

y 4

2 They would hug the asymptotes very closely!!!

0 -4 -2 0 2 4

-2

-4

So that’stwo votes for parabola and one vote for hyperbola. Majority rules. This surface is called a hyperbolic paraboloid. (b) This is clearly a function surface:

z = f (x; y) = x2 y2 or z = f (x; y) = y2 x2: The two 2-D parabolas meet at the origin, and that point (0; 0; 0) is called the saddle point. In one trace, it is a relative minimum, and in the other, it is a relative maximum. Thus, for the overall surface, it can never be an absolute minimum or maximum point.

9 (c) Notice that z = xy is also a hyperbolic paraboloid.

The entire surface has been rotated about the z-axis.

The rotational transformation requires a fair amount of knowledge. Sounds like a good extra credit assignment...

(#4) The Hyperboloids and the Elliptic Cone are essentially the same quadric surface.

(a) The hyperboloid of one sheet looks like this:

x2 + y2 z2 = 1: The standard form always has “1”on the right side alone.

We see that this particular surface is a surface of revolution about the z-axis. Thus, we expect the xz-plane trace to be identical to the yz-trace...

Also, all of the traces parallel to the xy-plane are circles, rather than ellipses. Below, we see the x = 0 and y = 0 traces. The cross-sections are identical.

z 4 z 4

3 3

2 2

1 1

0 0 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4

y x -1 -1

-2 -2

-3 -3

-4 -4 x2 y2 [Note that something like + z2 = 1 would NOT be a surface of revolution. 4 9 The traces parallel to the xy-plane would be proportional “2 by 3”ellipses, not circles.]

10 (b) The hyperboloid of two sheets looks like this:

x2 y2 + z2 = 1: The standard form always has “1”on the right side alone.

Again, we have chosen to display a surface of revolution about the z-axis.

For this particular surface, all of the traces parallel to the xy-plane are circles, rather than ellipses.

It should be clear that in the xz-plane and yz-plane traces, the hyperbolas open outward along the z-axis. I’mnot going to show those here. (c) The elliptic cones form the limiting boundary between the two previous cases. I could write the equation this way: x2 + y2 z2 = 0:

The standard form is z2 = x2 + y2:

It resembles the Pythagorean Theorem.

I have chosen to display right circular cones. The traces parallel to the xy-plane are circles. This is another surface of revolution.

Notice that we can obtain the upper cone only by taking the square root.

z = x2 + y2: p We could easily take the lower cone and move it upward 4 units:

z = x2 + y2 + 4: p We can create a solid cone through inequalities:

0 z x2 + y2 + 4: p [See next page.]

11 The base of this cone lies in the xy-plane. Its height is 4 units.

1 1 Its volume is R2h = 22 (4) = 3 3 16 cubic units. 3

(d) So how do we analyze the equations? (i) If it is the cones, then it always looks like the Pythagorean Theorem:

(something)2 (something)2 (something)2 = + : a2 b2 For example, we have elliptic cones:

(y)2 (x 3)2 = + (z 8)2 22 I can say that the vertex is located at V (3; 0; 8) : The central axis runs parallel to the x-axis because the lone term (on the left) has x in it. Thus, the cones open outward in the x-axis directions. This single cone opens outward in the negative-y direction:

x2 z2 y = + : s(0:42) (0:82)

12 (ii) If it is one of the hyperboloids, we must …rst obtain a standard form, with “1” on the right side alone. Count the negative signs on the left side. One negative sign = Hyperboloid of One Sheet. Two negative signs = Hyperboloid of Two Sheets. Back to the voting scheme... This time the variable vote with positive and negative signs! Whoever loses the vote (2 to 1) must tend the central axis. For example, we have a hyperboloid of one sheet (one negative sign in the standard form): z2 x2 + y2 + = 1: (0:3)2 The variables y and z have voted with positive signs and x loses this election with one negative sign. Thus, the central axis must be parallel to the x-axis. [See below, left.]

However, here we have a hyperboloid of two sheets (two negative signs in the standard form): y2 x2 + z2 = 1: (0:3)2 The variables x and z have voted with negative signs and y loses this election with one positive sign. Thus, the central axis must be parallel to the y-axis. [See above, right.]