#4: 1a) ANSWER: Hyperboloid of one sheet; r 2 ( 2 cos 2 0 + 3 sin 2
0 ) + r cos0 = 4z 2 + 5
SOLUTION: All three variables are squared. Two of the variables have the same signed coefficients
for their squared terms. Therefore, the surface is an hyperboloid. The center of the ellipse is x = -1/(2*2) = -1/4 and y = 0. Substituting these values into the equation results in - 4z 2 = 41/8 which has
no solution. The surface is an hyperboloid of one sheet.
Substituting the conversion equations:
2 ( r cos0 ) 2 + 3 ( r sin0 ) 2 - 4z 2 + ( r cos0 ) = 5
or r 2 ( 2 cos 2 0 + 3 sin 2 0 ) + r cos0 = 4z 2 + 5
#4: 1c) ANSWER: Plane; z = 8/3 - (r/3)( cos0 + 2 sin0 )
SOLUTION: The surface is a plane. Substituting the conversion equations: r cos0 +
2 r sin0 + 3z = 8
or z = 8/3 - (r/3)( cos0 + 2 sin0 )
#4: 1e) ANSWER: Cone; z = + r
SOLUTION: All three variables are squared. Two of the variables have the same signed coefficients
for their squared terms. Therefore, the surface is an hyperboloid. The center of the ellipse is x = 0 and y
= 0. Substituting these values into the equation results in - z 2 = 0 which has one solution. There fore
the surface is a cone. Substituting the conversion equations:
( r cos0 ) 2 + ( r sin0 ) 2 - z 2 =0 or r 2 = z 2 or z = + r
© 1999 Roxanne M. Byrne