Section 12.6 Cylinders and Quadric Surfaces
12. (a) Find and identify the traces of the quadric surface −x2− y2+ z2= 1 and explain why the graph looks like the graph of the hyperboloid of two sheets in Table 1.
(b) If the equation in part (a) is changed to x2− y2− z2= 1, what happens to the graph? Sketch the new graph.
Solution:
290 ¤ CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
10. (a) The traces of −2− 2+ 2= 1in = are −2+ 2 = 1 + 2, a family of hyperbolas, as are the traces in = ,
−2+ 2= 1 + 2. The traces in = are 2+ 2= 2− 1, a family of circles for || 1. As || increases, the radii of the circles increase; the traces are empty for || 1. This behavior, combined with the vertical traces, gives the graph of the hyperboloid of two sheets in Table 1.
(b) The graph has the same shape as the hyperboloid in part (a) but is rotated so that its axis is the -axis. Traces in = , || 1, are circles, while traces in = and = are hyperbolas.
11. 2= 2+ 42. The traces in = are the ellipses 2+ 42= 2. The traces in = are 2− 42= 2, hyperbolas for 6= 0 and two intersecting lines if = 0. Similarly, the traces in = are
2− 2= 42, hyperbolas for 6= 0 and two intersecting lines if = 0.
We recognize the graph as an elliptic cone with axis the -axis and vertex the origin.
12. 42+ 92+ 92= 36. The traces in = are 92+ 92= 36 − 42 ⇔
2+ 2= 4 −492, a family of circles for || 3. (The traces are a single point for || = 3 and are empty for || 3.) The traces in = are 42+ 92= 36 − 92, a family of ellipses for || 2. Similarly, the traces in = are the ellipses 42+ 92 = 36 − 92, || 2. The graph is an ellipsoid centered at the origin with intercepts = ±3, = ±2, = ±2.
13. 2= 42+ 2. The traces in = are the ellipses 42+ 2= 2. The traces in = are 2− 2= 42, hyperbolas for 6= 0 and two intersecting lines if = 0. Similarly, the traces in = are
2− 42= 2, hyperbolas for 6= 0 and two intersecting lines if = 0.
We recognize the graph as an elliptic cone with axis the -axis and vertex the origin.
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38. Reduce the equation to one of the standard forms, classify the surface, and sketch it. x2− y2− z2− 4x − 2z + 3 = 0.
Solution:
294 ¤ CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 33. 2+ 2 − 22= 0or 2 = 22− 2or = 2−2
2 represents a hyperbolic paraboloid with center (0 0 0).
34. 2= 2+ 42+ 4or −2+ 2− 42= 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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48. Find an equation for the surface obtained by rotating the line z = 2y about the z-axis.
Solution:
296 ¤ CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 42. We plot the surface = 2− 6 + 42.
43. 44.
45. The curve = √ is equivalent to = 2, ≥ 0. Rotating the curve about the -axis creates a circular paraboloid with vertex at the origin, axis the
-axis, opening in the positive -direction. The trace in the -plane is
= 2, = 0, and the trace in the -plane is a parabola of the same shape: = 2, = 0. An equation for the surface is = 2+ 2.
46. Rotating the line = 2 about the -axis creates a (right) circular cone with vertex at the origin and axis the -axis. Traces in = ( 6= 0) are circles with center (0 0 ) and radius = 2 = 2, so an equation for the trace is 2+ 2= (2)2, = . Thus an equation for the surface is
2+ 2= (2)2or 42+ 42= 2.
47. Let = (, , ) be an arbitrary point equidistant from (−1, 0, 0) and the plane = 1. Then the distance from to (−1, 0, 0) is
( + 1)2+ 2+ 2and the distance from to the plane = 1 is | − 1| √
12 = | − 1|
(by Equation 12.5.9). So | − 1| =
( + 1)2+ 2+ 2 ⇔ ( − 1)2 = ( + 1)2+ 2+ 2 ⇔
2− 2 + 1 = 2+ 2 + 1 + 2+ 2 ⇔ −4 = 2+ 2. Thus the collection of all such points is a circular paraboloid with vertex at the origin, axis the -axis, which opens in the negative -direction.
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50. Find an equation for the surface consisting of all points P for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface.
1
Solution:
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES ¤ 297
48. Let = ( ) be an arbitrary point whose distance from the -axis is twice its distance from the -plane. The distance from to the -axis is
( − )2+ 2+ 2=
2+ 2and the distance from to the -plane ( = 0) is || 1 = ||.
Thus
2+ 2= 2 || ⇔ 2+ 2= 42 ⇔ 2= (222) + (222). So the surface is a right circular cone with vertex the origin and axis the -axis.
49. (a) An equation for an ellipsoid centered at the origin with intercepts = ±, = ±, and = ± is 2
2 + 2
2 +2
2 = 1.
Here the poles of the model intersect the -axis at = ±6356523 and the equator intersects the - and -axes at
= ±6378137, = ±6378137, so an equation is
2
(6378137)2 + 2
(6378137)2 + 2
(6356523)2 = 1 (b) Traces in = are the circles 2
(6378137)2 + 2
(6378137)2 = 1 − 2
(6356523)2 ⇔
2+ 2= (6378137)2−
6378137 6356523
2
2.
(c) To identify the traces in = we substitute = into the equation of the ellipsoid:
2
(6378137)2 + ()2
(6378137)2 + 2
(6356523)2 = 1 (1 + 2)2
(6378137)2 + 2
(6356523)2 = 1
2
(6378137)2(1 + 2)+ 2
(6356523)2 = 1 As expected, this is a family of ellipses.
50. If we position the hyperboloid on coordinate axes so that it is centered at the origin with axis the -axis then its equation is given by 2
2 + 2
2 −2
2 = 1. Horizontal traces in = are 2
2 +2
2 = 1 +2
2, a family of ellipses, but we know that the traces are circles so we must have = . The trace in = 0 is 2
2 +2
2 = 1 ⇔ 2+ 2= 2and since the minimum radius of 100 m occurs there, we must have = 100. The base of the tower is the trace in = −500 given by
2
2 +2
2 = 1 + (−500)2
2 but = 100 so the trace is 2+ 2= 1002+ 50,0002 1
2. We know the base is a circle of radius 140, so we must have 1002+ 50,0002 1
2 = 1402 ⇒ 2= 50,0002
1402− 1002 = 781,250
3 and an equation for the tower is 2
1002 + 2
1002 − 2
(781,250)3= 1 or 2
10,000+ 2 10,000 −
32
781,250= 1, −500 ≤ ≤ 500.
51. If ( ) satisfies = 2− 2, then = 2− 2. 1: = + , = + , = + 2( − ),
2: = + , = − , = − 2( + ). Substitute the parametric equations of 1into the equation of the hyperbolic paraboloid in order to find the points of intersection: = 2− 2 ⇒
+ 2( − ) = ( + )2− ( + )2= 2− 2+ 2( − ) ⇒ = 2− 2. As this is true for all values of ,
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