FUNCTIONS OF SEVERAL VARIABLES
375
Describe the surface with the equation $ = k (0
41.88
41.89
41.90
41.91
41.92
41.93
41.94
41.95
41.96
41.97
41.98
<£ = k represents a one-napped cone with vertex at the origin whose generating lines make a fixed angle of k
radians with the positive z-axis.
Find a set of spherical coordinates for the point whose rectangular coordinates are (1,1, V6)
P
2 = 1
2 + 1
2 + (V6)
2 = 8. Hence, p = 2V2. tan <£ = Vl
2 + 1
2 /V6 = V2/V6 = 1/V3. Therefore, =
7T/6. tan 0={=1. Hence, 8 = ir/4. So, the spherical coordinates are (2V2, ir/6, ir/4
Find a set of spherical coordinates for the point whose rectangular coordinates are (0, -1, V5).
Hence,
7T/6. tan 0 =-1/0=-oo. Hence, 0 = 37r/2. So, a set of spherical coordinates is (2, rr/6, 37T/2).
Therefore, =
p = 2.
tan Find the rectangular coordinates of the point with spherical coordinates (3, -nil, ir/2).
Geometrically, it is easy to see that the point is (0,3,0). By calculation,
Find the rectangular coordinates of the point with spherical coordinates (4,2ir/3,17/3)
Find a spherical equation for the surface whose rectangular equation is x
2 + y
2 4- z
2 + 6z = 0.
x
2 + y
2 + z
2 = p
2
and
z = p cos .
Thus, we get p
2 +6p cos = 0. Then
p=0 or
p+
6cos=0. Since the origin is the only solution of p = 0 and the origin also lies on p + 6cos<£=0, then
p + 6 cos <{> — 0 is the desired equation.
Find a spherical equation for the surface whose rectangular equation is x + y = 4.
Of course, the surface is the right circular cylinder with radius 2 and the z-axis as axis of symmetry. Since
x
1 + y
2 = r
2 = p
2 sin
2 0, we have p
2 sin
2 $=4. This can be reduced to p sin = 2. (The other possibility
p sin <£ = ~2 yields no additional points.)
Find the graph of the spherical equation p = 2a sin, where a > 0.
Since 9 is not present in the equation, the surface is obtained by rotating about the z-axis the intersection of the
surface with the yz-plane. In the latter plane, p=2asin<£ becomes
2a>>, (y - a) +z =a. This is a circle with center (o,0) and radius a. So, the resulting surf ace is obtained by
rotating that circle about the z-axis. The result can be thought of as a doughnut (torus) with no hole in the
middle.
Describe the surface whose equation in spherical coordinates is p sin (f> = 3.
Since r = p sin = 3, this is the right circular cylinder with radius 3 and the z-axis as axis of symmetry.
Describe the surface whose equation in spherical coordinates is p cos <£ = 3.
Since z = p cos <£ = 3, this is the plane that is parallel to, and three units above, the ry-plane.
Describe the surface whose equation in spherical coordinates is p
2 sin
2 cos 26 = 4.
p
2 sin
2 0 cos 20 = p
2 sin
2 (cos
2 0 - sin
2 0) = p
2 sin
2 $ cos
2 0 - p
2 sin
2 sin
2 6 = x
2 - y
2
. Hence, we have
the cylindrical surface x
2 - y
2 = 4, generated by the hyperbola x
2 - y
2 = 4 in the jcy-plane.
Find an equation in spherical coordinates of the ellipsoid x
2 + y
2 + 9z
2 = 9.
x
2 + y
2 + 9z
2 = jc
2 + y
2 + z
2 + 8z
2 = p
2 + 8p
2 cos
2 $ = 9. Thus, we obtain the equation p
2 (l + 8 cos
2 0) = 9.
p
2 = 0
2 + (-l)
2 + (V3)
2 = 4.
/+**=
375
Describe the surface with the equation $ = k (0
41.88
41.89
41.90
41.91
41.92
41.93
41.94
41.95
41.96
41.97
41.98
<£ = k represents a one-napped cone with vertex at the origin whose generating lines make a fixed angle of k
radians with the positive z-axis.
Find a set of spherical coordinates for the point whose rectangular coordinates are (1,1, V6)
P
2 = 1
2 + 1
2 + (V6)
2 = 8. Hence, p = 2V2. tan <£ = Vl
2 + 1
2 /V6 = V2/V6 = 1/V3. Therefore,
7T/6. tan 0={=1. Hence, 8 = ir/4. So, the spherical coordinates are (2V2, ir/6, ir/4
Find a set of spherical coordinates for the point whose rectangular coordinates are (0, -1, V5).
Hence,
7T/6. tan 0 =-1/0=-oo. Hence, 0 = 37r/2. So, a set of spherical coordinates is (2, rr/6, 37T/2).
Therefore,
p = 2.
tan Find the rectangular coordinates of the point with spherical coordinates (3, -nil, ir/2).
Geometrically, it is easy to see that the point is (0,3,0). By calculation,
Find the rectangular coordinates of the point with spherical coordinates (4,2ir/3,17/3)
Find a spherical equation for the surface whose rectangular equation is x
2 + y
2 4- z
2 + 6z = 0.
x
2 + y
2 + z
2 = p
2
and
z = p cos .
Thus, we get p
2 +6p cos
p=0 or
p+
6cos=0. Since the origin is the only solution of p = 0 and the origin also lies on p + 6cos<£=0, then
p + 6 cos <{> — 0 is the desired equation.
Find a spherical equation for the surface whose rectangular equation is x + y = 4.
Of course, the surface is the right circular cylinder with radius 2 and the z-axis as axis of symmetry. Since
x
1 + y
2 = r
2 = p
2 sin
2 0, we have p
2 sin
2 $=4. This can be reduced to p sin
p sin <£ = ~2 yields no additional points.)
Find the graph of the spherical equation p = 2a sin
Since 9 is not present in the equation, the surface is obtained by rotating about the z-axis the intersection of the
surface with the yz-plane. In the latter plane, p=2asin<£ becomes
2a>>, (y - a) +z =a. This is a circle with center (o,0) and radius a. So, the resulting surf ace is obtained by
rotating that circle about the z-axis. The result can be thought of as a doughnut (torus) with no hole in the
middle.
Describe the surface whose equation in spherical coordinates is p sin (f> = 3.
Since r = p sin
Describe the surface whose equation in spherical coordinates is p cos <£ = 3.
Since z = p cos <£ = 3, this is the plane that is parallel to, and three units above, the ry-plane.
Describe the surface whose equation in spherical coordinates is p
2 sin
2
p
2 sin
2 0 cos 20 = p
2 sin
2
2 0 - sin
2 0) = p
2 sin
2 $ cos
2 0 - p
2 sin
2
2 6 = x
2 - y
2
. Hence, we have
the cylindrical surface x
2 - y
2 = 4, generated by the hyperbola x
2 - y
2 = 4 in the jcy-plane.
Find an equation in spherical coordinates of the ellipsoid x
2 + y
2 + 9z
2 = 9.
x
2 + y
2 + 9z
2 = jc
2 + y
2 + z
2 + 8z
2 = p
2 + 8p
2 cos
2 $ = 9. Thus, we obtain the equation p
2 (l + 8 cos
2 0) = 9.
p
2 = 0
2 + (-l)
2 + (V3)
2 = 4.
/+**=