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(b1) (5,−5π/6, π/3) is already in the canonical set.
(b2) x = r cos p sin h = 5 cos(π/3) sin(−5π/6) = 5(1/2)(−1/2) = −5/4
y = −r sin p = −5 sin(π/3) = −5(
3/2) = −(5
3)/2
z = r cos p cos h = 5 cos(π/3) cos(−5π/6) = 5(1/2)(−
3/2) = −(5
3)/4
3)/4)
so (x, y, z) = (−5/4,−(5
3)/2,−(5
(c1) (2,−π/6, π) =⇒ (2, 5π/6, 0)
(c2) x = r cos p sin h = 2 cos(0) sin(5π/6) = (2)(1)(1/2) = 1
y = −r sin p = −2 sin(0) = (−2)(0) = 0
z = r cos p cos h = 2 cos(0) cos(5π/6) = (2)(1)(−
3/2) = −
3
so (x, y, z) = (1, 0,−
3)
(d1) (8, 9π/4, π/6) =⇒ (8, π/4, π/6)
(d2) x = r cos p sin h = 8 cos(π/6) sin(π/4) = 8(
2/2) = 2
6
3/2)(
y = −r sin p = −8 sin(π/6) = −8(1/2) = −4
z = r cos p cos h = 8 cos(π/6) cos(π/4) = 8(
3/2)(
2/2) = 2
6
so (x, y, z) = (2
6,−4, 2
6)
3) 2 + (−
16 = 4
10.
(a) r =
x 2 + y 2 + z 2 =
(
2) 2 + (2
2) 2 =
2 + 12 + 2 =
h = arctan(x/z) = arctan(−
2) = arctan(−1) = 135 o , given the location of
2/
(x, z)
p = arcsin(−y/r) = arcsin(−(2
3)/4) = arcsin(−
3/2) = −60 o
so (r, h, p) = (4, 135 o ,−60 o )
3) 2 + 6 2 + (−4) 2 =
(b) r =
x 2 + y 2 + z 2 =
(2
12 + 36 + 16 =
64 = 8
3)/4) = arctan(−
3/2) = 139.11 o , given the loca-
h = arctan(x/z) = arctan(−(2
tion of (x, z)
p = arcsin(−y/r) = arcsin(−6/8) = arcsin(−3/4) = −48.59 o
so (r, h, p) = (8, 139.11 o ,−48.59 o )
(c) r =
1 + 1 + 1 =
3
h = arctan(x/z) = arctan((−1)/(−1)) = arctan(1) = −135 o , given the location of
(x, z)
p = arcsin(−y/r) = arcsin(1/
x 2 + y 2 + z 2 =
(−1) 2 + (−1) 2 + (−1) 2 =
3) = 35.26 o
3,−135 o , 35.26 o )
so (r, h, p) = (
2
h = arctan(x/z) = arctan(2/4) = arctan(1/2) = 26.57 o , given the location of (x, z)
p = arcsin(−y/r) = arcsin((2
32 = 4
(d) r =
x 2 + y 2 + z 2 =
2 2 + (−2
3) 2 + 4 2 =
4 + 12 + 16 =
3)/(4
2)) = 37.76 o
2)) = arcsin(
3/(2
2, 26.57 o , 37.76 o )
so (r, h, p) = (4
(−
3) 2 + (−
2) 2 =
(e) r =
x 2 + y 2 + z 2 =
3) 2 + (2
3 + 3 + 8 =
14
h = arctan(x/z) = arctan(−
2)) = −31.48 o , given the location of (x, z)
3/(2
3/
14) = 27.58 o
p = arcsin(−y/r) = arcsin(
14,−31.48 o , 27.58 o )
so (r, h, p) = (
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