Section 12.6 Cylinders and Quadric Surfaces

Section 12.6 Cylinders and Quadric Surfaces

SECTION 12.6 CYLINDERS AND QUADRIC SURFACES ¤ 289

7.Since  is missing, each horizontal trace  = 1,

 = , is a copy of the same hyperbola in the plane

 = . Thus the surface  = 1 is a hyperbolic cylinder with rulings parallel to the -axis.

8.Since  is missing, each vertical trace  = sin ,

 = , is a copy of a sine curve in the plane  = .

Thus the surface  = sin  is a cylindrical surface with rulings parallel to the -axis.

9. (a) The traces of ²+ ²− ²= 1in  =  are ²− ²= 1 − ², a family of hyperbolas. (Note that the hyperbolas are oriented differently for −1    1 than for   −1 or   1.) The traces in  =  are ²− ²= 1 − ², a similar family of hyperbolas. The traces in  =  are ²+ ²= 1 + ², a family of circles. For  = 0, the trace in the

-plane, the circle is of radius 1. As || increases, so does the radius of the circle. This behavior, combined with the hyperbolic vertical traces, gives the graph of the hyperboloid of one sheet in Table 1.

(b) The shape of the surface is unchanged, but the hyperboloid is rotated so that its axis is the -axis. Traces in  =  are circles, while traces in  =  and  =  are hyperbolas.

(c) Completing the square in  gives ²+ ( + 1)²− ²= 1. The surface is a hyperboloid identical to the one in part (a) but shifted one unit in the negative -direction.

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292 ¤ CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE

18. 3²− ²+ 3²= 0. The traces in  =  are ²− 3²= 3², a family of hyperbolas for  6= 0 and two intersecting lines if  = 0. Traces in  =  are the circles 3²+ 3²= ² ⇔ ²+ ²=¹₃². The traces in  =  are ²− 3²= 3², hyperbolas for  6= 0 and two intersecting lines if

 = 0. We recognize the surface as a circular cone with axis the -axis and vertex the origin.

19.  = ²− ². The traces in  =  are the parabolas  = ²− ², opening in the positive -direction. The traces in  =  are  = ²− ², two intersecting lines when  = 0 and a family of hyperbolas for  6= 0 (note that the hyperbolas are oriented differently for   0 than for   0). The traces in  =  are the parabolas  = ²− ²which open in the negative

-direction. Thus the surface is a hyperbolic paraboloid centered at (0 0 0).

20.  = ²− ². The traces in  =  are ²− ²= , two intersecting lines when  = 0 and a family of hyperbolas for  6= 0 (oriented differently for

  0than for   0). The traces in  =  are the parabolas

 = −²+ ², opening in the negative -direction, and the traces in  =  are the parabolas  = ²− ²which open in the positive -direction. The graph is a hyperbolic paraboloid centered at (0 0 0).

21. This is the equation of an ellipsoid: ²+ 4²+ 9²= ²+ ²

(12)² + ²

(13)² = 1, with -intercepts ±1, -intercepts ±¹2

and -intercepts ±¹3. So the major axis is the -axis and the only possible graph is VII.

22. This is the equation of an ellipsoid: 9²+ 4²+ ²= ²

(13)² + ²

(12)² + ² = 1, with -intercepts ±¹3, -intercepts ±¹2

and -intercepts ±1. So the major axis is the -axis and the only possible graph is IV.

23. This is the equation of a hyperboloid of one sheet, with  =  =  = 1. Since the coefficient of ²is negative, the axis of the hyperboloid is the -axis, hence the correct graph is II.

24. This is a hyperboloid of two sheets, with  =  =  = 1. This surface does not intersect the -plane at all, so the axis of the hyperboloid is the -axis and the graph is III.

25. There are no real values of  and  that satisfy this equation for   0, so this surface does not extend to the left of the

-plane. The surface intersects the plane  =   0 in an ellipse. Notice that  occurs to the first power whereas  and  occur to the second power. So the surface is an elliptic paraboloid with axis the -axis. Its graph is VI.

26. This is the equation of a cone with axis the -axis, so the graph is I.

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292 ¤ CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE

18. 3²− ²+ 3²= 0. The traces in  =  are ²− 3²= 3², a family of hyperbolas for  6= 0 and two intersecting lines if  = 0. Traces in  =  are the circles 3²+ 3²= ² ⇔ ²+ ²=¹₃². The traces in  =  are ²− 3²= 3², hyperbolas for  6= 0 and two intersecting lines if

 = 0. We recognize the surface as a circular cone with axis the -axis and vertex the origin.

19.  = ²− ². The traces in  =  are the parabolas  = ²− ², opening in the positive -direction. The traces in  =  are  = ²− ², two intersecting lines when  = 0 and a family of hyperbolas for  6= 0 (note that the hyperbolas are oriented differently for   0 than for   0). The traces in  =  are the parabolas  = ²− ²which open in the negative

-direction. Thus the surface is a hyperbolic paraboloid centered at (0 0 0).

20.  = ²− ². The traces in  =  are ²− ²= , two intersecting lines when  = 0 and a family of hyperbolas for  6= 0 (oriented differently for

  0than for   0). The traces in  =  are the parabolas

 = −²+ ², opening in the negative -direction, and the traces in  =  are the parabolas  = ²− ²which open in the positive -direction. The graph is a hyperbolic paraboloid centered at (0 0 0).

21. This is the equation of an ellipsoid: ²+ 4²+ 9²= ²+ ²

(12)² + ²

(13)² = 1, with -intercepts ±1, -intercepts ±¹2

and -intercepts ±¹3. So the major axis is the -axis and the only possible graph is VII.

22. This is the equation of an ellipsoid: 9²+ 4²+ ²= ²

(13)² + ²

(12)² + ² = 1, with -intercepts ±¹3, -intercepts ±¹2

and -intercepts ±1. So the major axis is the -axis and the only possible graph is IV.

23. This is the equation of a hyperboloid of one sheet, with  =  =  = 1. Since the coefficient of ²is negative, the axis of the hyperboloid is the -axis, hence the correct graph is II.

24. This is a hyperboloid of two sheets, with  =  =  = 1. This surface does not intersect the -plane at all, so the axis of the hyperboloid is the -axis and the graph is III.

25. There are no real values of  and  that satisfy this equation for   0, so this surface does not extend to the left of the

-plane. The surface intersects the plane  =   0 in an ellipse. Notice that  occurs to the first power whereas  and  occur to the second power. So the surface is an elliptic paraboloid with axis the -axis. Its graph is VI.

26. This is the equation of a cone with axis the -axis, so the graph is I.

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SECTION 12.6 CYLINDERS AND QUADRIC SURFACES ¤ 293

27. This surface is a cylinder because the variable  is missing from the equation. The intersection of the surface and the -plane is an ellipse. So the graph is VIII.

28. This is the equation of a hyperbolic paraboloid. The trace in the -plane is the parabola  = ². So the correct graph is V.

29. Vertical traces parallel to the -plane are circles centered at the origin whose radii increase as  decreases. (The trace in

 = 1is just a single point and the graph suggests that traces in  =  are empty for   1.) The traces in vertical planes parallel to the -plane are parabolas opening to the left that shift to the left as || increases. One surface that fits this description is a circular paraboloid, opening to the left, with vertex (0 1 0).

30. The vertical traces parallel to the -plane are ellipses that are smallest in the -plane and increase in size as || increases. One surface that fits this description is a hyperboloid of one sheet with axis the -axis. The horizontal traces in  =  (hyperbolas and intersecting lines) also fit this surface, as shown in the figure.

31. ²= ²+¹₉²or ²= ²+²

9 represents an elliptic cone with vertex (0 0 0) and axis the -axis.

32.4²−  + 2²= 0or  = ² 14+ ²

12or 

4= ²+² 2 represents an elliptic paraboloid with vertex (0 0 0) and axis the -axis.

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294 ¤ CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 33. ²+ 2 − 2²= 0or 2 = 2²− ²or  = ²−²

2 represents a hyperbolic paraboloid with center (0 0 0).

34.²= ²+ 4²+ 4or −²+ ²− 4²= 4or

−² 4 +²

4 − ²= 1represents a hyperboloid of two sheets with axis the -axis.

35. Completing squares in  and  gives

²− 2 + 1 +

²− 6 + 9

−  = 0 ⇔

( − 1)²+ ( − 3)²−  = 0 or  = ( − 1)²+ ( − 3)², a circular paraboloid opening upward with vertex (1 3 0) and axis the vertical line

 = 1,  = 3.

36. Completing squares in  and  gives

²− 4 + 4

− ²−

²+ 2 + 1

+ 3 = 0 + 4 − 1 ⇔ ( − 2)²− ²− ( + 1)² = 0or ( − 2)²= ²+ ( + 1)², a circular

cone with vertex (2 0 −1) and axis the horizontal line  = 0,  = −1.

37. Completing squares in  and  gives

²− 4 + 4

− ²+

²− 2 + 1

= 0 + 4 + 1 ⇔

( − 2)²− ²+ ( − 1)² = 5or ( − 2)²

5 −²

5 +( − 1)² 5 = 1, a hyperboloid of one sheet with center (2 0 1) and axis the horizontal line

 = 2,  = 1.

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