This preview shows page 1. Sign up to view the full content.
Unformatted text preview: has radius 1.
1. x2 + z 2 − 4x = 0
2. x2 + y 2 + 4x = 0
3. z 2 + x2 + 2z = 0 correct
4. z 2 + x2 − 2x = 0 keywords: quadric surface, graph of equation,
cylinder, 3D graph, circular cylinder, trace
014 5. x2 + y 2 + 2x = 0
2 as a circle in the xz plane.
Consequently, the graph is that of the equation
z 2 + x2 + 2 z = 0 . 10.0 points Reduce the equation 2 6. z + x + 4z = 0 x2 − y 2 + z 2 + 2 x + 2 y + 4 z + 2 = 0 Explanation:
The graph is a circular cylinder whose axis
of symmetry is parallel to the y axis, so it
will be the graph of an equation containing no
y term. This already eliminates the equations to standard form and then classify the surface.
(x + 1)2 (y − 1)2 (z + 2)2
+
+
= 1,
1.
2
2
2
ellipsoid
(x + 1)2 (y + 1)2 (z − 2)2
2.
−
+
= 1,
2
2
2
hyperboloid x2 + y 2 + 2 x = 0 , x2 + y 2 + 4 x = 0 .
(x + 1)2 (y + 1)2 (z + 2)2
−
+
= 1,
3.
2
2
2
hyperboloid On the other hand, the intersection of the
graph with the xz plane, i.e. the y = 0 plane,
is a circle centered on the z axis and passing
through the origin as shown in (x − 1)2 (y − 1)2 (z + 2)2
+
+
= 1,
4.
2
2
2
ellipsoid geyer (dag2798) – HW09 – rusin – (55735)
(x + 1)2 (y − 1)2 (z + 2)2
5.
−
+
= 1,
2
2
2
hyperboloid correct
Explanation:
Completing squares in all three variables
gives (x + 1)2 − (y − 1)2 + (z + 2...
View
Full
Document
This note was uploaded on 05/12/2013 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.
 Spring '07
 Sadler
 Determinant, Multivariable Calculus

Click to edit the document details