product. To be specific, if
a1=(a11,a12, . . . ,a1n), a2=(a21,a22, . . . ,a2n), . . . , an−1=(an−11,an−12, . . . ,an−1n)
aren−1vectors inRn, we definea1×a2×· · ·×an−1to be the vector inRngiven by the symbolic determinant
a1×a2×· · ·×an−1=
e1 e2 · · · en
a11 a12 · · · a1n
a21 a22 · · · a2n
... ... . .. ... an−1 1 an−1 2 · · · an−1n
.
(Heree1, . . . ,enare the standard basis vectors forRn.)Exer- cises 39–42 concern this generalized notion of cross product.
39. Calculate the following cross product inR4: (1,2,−1,3)×(0,2,−3,1)×(−5,1,6,0).
40. Use the results of Exercises 26 and 28 to show that (a) a1× · · · ×ai× · · · ×aj× · · · ×an−1
= −(a1× · · · ×aj× · · · ×ai× · · · ×an−1), 1≤i≤n−1,1≤ j ≤n−1
(b) a1× · · · ×kai× · · · ×an−1
=k(a1× · · · ×ai× · · · ×an−1), 1≤i≤n−1.
(c) a1× · · · ×(ai+b)× · · · ×an−1
=a1× · · · ×ai× · · · ×an−1+ a1× · · · × b× · · · ×an−1, 1≤i≤n−1, allb∈Rn.
(d) Show that ifb=(b1, . . . ,bn) is any vector inRn, then
b·(a1×a2×· · ·×an−1) is given by the determinant
b1 · · · bn
a11 · · · a1n
... ...
an−11 · · · an−1n
.
41. Show that the vector b=a1×a2×· · ·×an−1 is orthogonal toa1, . . . ,an−1.
42. Use the generalized notion of cross products to find an equation of the (four-dimensional) hyper- plane inR5 through the five points P0(1,0,3,0,4), P1(2,−1,0,0,5), P2(7,0,0,2,0), P3(2,0,3,0,4), andP4(1,−1,3,0,4).
P
Figure 1.84 The polar coordinate system.
P r
θ
Figure 1.85 Locating a pointP, using polar coordinates.
exactly one such circle and one such ray. The origin itself is special: No circle passes through it, and all the rays begin at it. (See Figure 1.84.) For pointsPother than the origin, we assign toPthe polar coordinates (r, θ), wherer is the radius of the circle on which Plies andθ is the angle between the positivex-axis and the ray on whichPlies. (θis measured as opening counterclockwise.) The origin is an exception: It is assigned the polar coordinates (0, θ), where θ can be any angle. (See Figure 1.85.) As we have described polar coordinates,r ≥0 sincer is the radius of a circle. It also makes good sense to require 0≤θ <2π, for then every point in the plane, except the origin, has a uniquely determined pair of polar coordinates. Occasionally, however, it is usefulnotto restrictrto be nonnegative andθto be between 0 and 2π. In such a case,nopoint ofR2will be described by a unique pair of polar coordinates: If Phas polar coordinates (r, θ), then it also has (r, θ+2nπ) and (−r, θ+(2n+1)π) as coordinates, where n can be any integer. (To locate the point having coordinates (r, θ), wherer <0, construct the ray making angleθ with respect to the positivex-axis, and instead of marching
|r|units away from the origin along this ray, go|r|units in theoppositedirection, as shown in Figure 1.86.)
(−|r|, )θ
(|r|, )θ θ
Figure 1.86 Locating the point with polar coordinates (r, θ), wherer<0.
EXAMPLE 1 Polar coordinates may already be familiar to you. Nonetheless, make sure you understand that the points pictured in Figure 1.87 have the coor-
dinates indicated. ◆
(5, 0) (2, 5 /6)π
(−1, 5 /6) orπ (1, 11 /6) orπ (1, − π/6) (3, 3 /2)π
(2, /6)π
Figure 1.87 Figure for Example 1.
(3√⎯ π2, /4)
(3√⎯ π3, /6)
(6, 0)
Figure 1.88 The graph of r=6 cosθin Example 2.
θ r=6 cosθ
0 6
π/6 3√
3
π/4 3√
2
π/3 3
π/2 0
2π/3 −3
3π/4 −3√
2
5π/6 −3√
3
π −6
7π/6 −3√
3
5π/4 −3√
2 4π/3 −3
3π/2 0
5π/3 3
7π/4 3√ 2
EXAMPLE 2 Let’s graph the curve given by the polar equation r =6 cosθ (Figure 1.88). We can begin to get a feeling for the graph by compiling values, as in the adjacent tabulation.
Thus,rdecreases from 6 to 0 asθincreases from 0 toπ/2;rdecreases from 0 to−6 (or is not defined, if you taker to be nonnegative) asθ varies fromπ/2 toπ;r increases from−6 to 0 asθ varies fromπ to 3π/2; andr increases from 0 to 6 asθvaries from 3π/2 to 2π. To graph the resulting curve, imagine a radar screen: Asθmoves counterclockwise from 0 to 2π, the point (r, θ) of the graph is traced as the appropriate “blip” on the radar screen. Note that the curve is actually traced twice: once asθ varies from 0 toπ and then again asθ varies fromπ to 2π. Alternatively, the curve is traced just once if we allow onlyθvalues that yield nonnegativer values. The resulting graph appears to be a circle of radius 3 (not centered at the origin), and, in fact, one can see (as in Example 3) that the graph
is indeed such a circle. ◆
The basic conversions between polar and Cartesian coordinates are provided by the following relations:
Polar to Cartesian:
x =rcosθ
y =rsinθ ; (1)
Cartesian to polar:
r2=x2+y2
tanθ = y/x . (2)
Note that the equations in (2) do not uniquely determiner andθ in terms of x andy. This is quite acceptable, really, since we do not always want to insist that r be nonnegative andθ be between 0 and 2π. If we do restrictrandθ, however, then they are given in terms ofx andyby the following formulas:
r =
x2+y2,
θ=
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
tan−1y/x ifx>0,y≥0 tan−1y/x+2π ifx>0,y<0 tan−1y/x+π ifx<0,y≥0
π/2 ifx=0,y>0
3π/2 ifx=0,y<0 indeterminate ifx= y=0
.
The complicated formula forθarises because we require 0≤θ <2π, while the inverse tangent function returns values between−π/2 andπ/2 only. Now you see why the equations given in (2) are a better bet!
EXAMPLE 3 We can use the formulas in (1) and (2) to prove that the curve in Example 2 really is a circle. The polar equationr =6 cosθthat defines the curve requires a little ingenuity to convert to the corresponding Cartesian equation. The trick is to multiply both sides of the equation byr. Doing so, we obtain
r2 =6rcosθ.
Now (1) and (2) immediately give
x2+y2 =6x.
We complete the square inxto find that this equation can be rewritten as (x−3)2+y2 =9,
which is indeed a circle of radius 3 with center at (3,0). ◆ y
x
z
Figure 1.89 The cylindrical coordinate system.
z θ r
θ P(r, , z)
Figure 1.90 Locating a pointP, using cylindrical coordinates.
Cylindrical Coordinates
Cylindrical coordinates onR3are a “naive” way of generalizing polar coordinates to three dimensions, in the sense that they are nothing more than polar coordinates used in place of thex- andy-coordinates. (Thez-coordinate is left unchanged.) The geometry is as follows: Except for thez-axis, fill all of space with infinitely extended circular cylinders with axes along thez-axis as in Figure 1.89. Then any point P in R3 not lying on the z-axis lies on exactly one such cylinder.
Hence, to locate such a point, it’s enough to give the radius of the cylinder, the circumferential angleθ around the cylinder, and the vertical positionzalong the cylinder. Thecylindrical coordinatesofPare (r, θ,z), as shown in Figure 1.90.
Algebraically, the equations in (1) and (2) can be extended to produce the basic conversions between Cartesian and cylindrical coordinates.
The basic conversions between cylindrical and Cartesian coordinates are provided by the following relations:
Cylindrical to Cartesian:
⎧⎪
⎨
⎪⎩
x =rcosθ y=rsinθ z=z
; (3)
Cartesian to cylindrical:
⎧⎪
⎨
⎪⎩
r2 =x2+y2 tanθ= y/x z=z
. (4)
r0
Figure 1.91 The graph of the cylindrical equationr=r0.
As with polar coordinates, if we make the restrictionsr ≥0, 0≤θ <2π, then all points ofR3 except thez-axis have a unique set of cylindrical coordinates. A point on thez-axis with Cartesian coordinates (0,0,z0) has cylindrical coordinates (0, θ,z0), whereθcan be any angle.
Cylindrical coordinates are useful for studying objects possessing an axis of symmetry. Before exploring a few examples, let’s understand the three “constant coordinate” surfaces.
• Ther =r0 surface is, of course, just a cylinder of radiusr0 with axis the z-axis. (See Figure 1.91.)
• Theθ =θ0surface is a vertical plane containing thez-axis (or a half-plane with edge thez-axis if we taker ≥0 only). (See Figure 1.92.)
• Thez =z0surface is a horizontal plane. (See Figure 1.93.)
Half-plane only if r ≥ 0
θ0
Figure 1.92 The graph ofθ =θ0.
z = z0
Figure 1.93 The graph of z=z0.
x y
z
z = c
Figure 1.94 The graph of r=6 cosθin cylindrical coordinates.
EXAMPLE 4 Graph the surface having cylindrical equationr =6 cosθ. (This equation is identical to the one in Example 2.) In particular,z does not appear in this equation. What this means is that if the surface is sliced by the horizontal planez=cwherecis a constant, we will see the circle shown in Example 2, no matter whatcis. If we stack these circular sections, then the entire surface is a circular cylinder of radius 3 with axis parallel to thez-axis (and through the point (3,0,0) in cylindrical coordinates). This surface is shown in Figure 1.94. ◆ EXAMPLE 5 Graph the surface having equation z=2r in cylindrical co- ordinates.
Here the variableθ does not appear in the equation, which means that the surface in question will be circularly symmetric about thez-axis. In other words, if we slice the surface by any plane of the formθ =constant (or half-plane, if we taker ≥0), we see the same curve, namely, a line (respectively, a half-line) of slope 2. As we let the constant-θ plane vary, this line generates a cone, as shown in Figure 1.95. The cone consists only of the top half (nappe) when we restrictr to be nonnegative.
= c θ z
y x
Figure 1.95 The graph ofz=2r in cylindrical coordinates.
The Cartesian equation of this cone is readily determined. Using the formulas in (4), we have
z =2r =⇒ z2=4r2 ⇐⇒ z2=4(x2+y2).
Sincezcan be positive as well as negative, this last Cartesian equation describes the cone with both nappes. If we want the top nappe only, then the equation z=2
x2+y2 describes it. Similarly, z= −2
x2+y2 describes the bottom
nappe. ◆
Spherical Coordinates
Fill all of space with spheres centered at the origin as in Figure 1.96. Then every pointP ∈R3, except the origin, lies on a single such sphere. Roughly speaking, thespherical coordinatesofPare given by specifying the radiusρof the sphere containing P and the “latitude and longitude” readings of P along this sphere.
More precisely, the spherical coordinates (ρ, ϕ, θ) ofPare defined as follows:ρ is the distance fromPto the origin;ϕis the angle between the positivez-axis and the ray through the origin and P; andθ is the angle between the positivex-axis and the ray made by dropping a perpendicular fromPto thex y-plane. (See Figure 1.97.) Theθ-coordinate is exactly the same as theθ-coordinate used in cylindrical coordinates. (Warning:Physicists usually prefer to reverse the roles ofϕandθ, as do some graphical software packages.)
z
y
x
Figure 1.96 The spherical coordinate system.
P ρ ϕ
θ
Figure 1.97 Locating the point P, using spherical coordinates.
It is standard practice to impose the following restrictions on the range of values for the individual coordinates:
ρ≥0, 0≤ϕ≤π, 0≤θ <2π. (5)
With such restrictions, all points ofR3, except those on thez-axis, have a uniquely determined set of spherical coordinates. Points along thez-axis, except for the origin, have coordinates of the form (ρ0,0, θ) or (ρ0, π, θ), whereρ0is a positive constant andθ is arbitrary. The origin has spherical coordinates (0, ϕ, θ), where bothϕandθare arbitrary.
EXAMPLE 6 Several points and their corresponding spherical coordinates are
shown in Figure 1.98. ◆
/4 (2, /4, /2)π π
(1, /4, 0)π
(2, /2, /2)π π
(2, , /4) π π or (2, , /3) or (−2, 0, 0)π π π
Figure 1.98 Figure for Example 6. Figure 1.99 The graph of ρ=ρ0(>0).
ϕ0< π/2
ϕ0 = /2π
ϕ0 > /2π
Figure 1.100 The spherical surfaceϕ=ϕ0, shown for different values ofϕ0.
Spherical coordinates are especially useful for describing objects that have a center of symmetry. With the restrictions given by the inequalities in (5), the constant coordinate surfaceρ=ρ0(ρ0 >0) is, of course, a sphere of radiusρ0, as shown in Figure 1.99. The surface given byθ =θ0is a half-plane just as in the cylindrical case. Theϕ=ϕ0 surface is a single-nappe cone ifϕ0=π/2 and is thex y-plane ifϕ0 =π/2 (and is the positive or negativez-axis ifϕ0 =0 orπ).
(See Figure 1.100.) If we do not insist thatρbe nonnegative, then the cones would include both nappes.
The basic equations relating spherical coordinates to both cylindrical and Cartesian coordinates are as follows.
Spherical/cylindrical:
⎧⎨
⎩
r =ρsinϕ θ=θ z=ρcosϕ
⎧⎨
⎩
ρ2 =r2+z2 tanϕ=r/z
θ=θ . (6)
Spherical/Cartesian:
⎧⎨
⎩
x=ρsinϕ cosθ y=ρsinϕ sinθ z=ρcosϕ
⎧⎨
⎩
ρ2 =x2+y2+z2 tanϕ=
x2+y2/z tanθ = y/x
. (7)
θ
z ρ
r ϕ
ϕ /2 − π
Figure 1.101 Converting spherical to cylindrical coordinates when 0< ϕ <π2.
ρ r
z(< 0) − /2π
ϕ
ϕ
Figure 1.102 Converting spherical to cylindrical
coordinates whenπ/2< ϕ < π. Using basic trigonometry, it is not difficult to establish the conversions in (6).
From the right triangle shown in Figure 1.101, we have cos
π 2 −ϕ
= r ρ. Hence,
r =ρcos π
2 −ϕ
=ρsinϕ.
Similarly,
sin π
2 −ϕ
= z ρ, so that
z=ρsin π
2 −ϕ
=ρcosϕ.
Thus, the formulas in (6) follow when 0≤ϕ≤π/2. Ifπ/2< ϕ≤π, then we may employ Figure 1.102. So
r =ρcos ϕ−π
2
=ρsinϕ, and
z= −ρsin
ϕ− π 2
=ρsin π
2 −ϕ
=ρcosϕ.
Hence, the relations in (6) hold in general. The equations in (7) follow by substi- tution of those in (6) into those of (3) and (4).
EXAMPLE 7 The cylindrical equation z=2r in Example 5 converts via (6) to the spherical equation
ρcosϕ=2ρsinϕ.
Therefore,
tanϕ= 1
2 ⇐⇒ ϕ=tan−1 1 2 ≈26◦.
Thus, the equation defines a cone (as we just saw). The spherical equation is especially simple in that it involves just a single coordinate. ◆ EXAMPLE 8 Not all spherical equations are improvements over their cylindri- cal or Cartesian counterparts. For example, the Cartesian equation 6x =x2+y2 (whose polar–cylindrical equivalent isr =6 cosθ) becomes
6ρsinϕcosθ =ρ2sin2ϕ cos2θ+ρ2sin2ϕsin2θ from (7). Simplifying,
6ρsinϕ cosθ =ρ2sin2ϕ(cos2θ+sin2θ)
⇐⇒ 6ρsinϕ cosθ =ρ2sin2ϕ
⇐⇒ 6 cosθ =ρsinϕ.
This spherical equation ismorecomplicated than the original Cartesian equation in that all three spherical coordinates are involved. Therefore, it is not at all obvious that the spherical equation describes a cylinder. ◆ ϕ ρ=2acosϕ
0 2a
π/6 √
3a
π/4 √
2a
π/3 a
π/2 0
2π/3 −a
3π/4 −√
π −2a2a
EXAMPLE 9 Let’s graph the surface with spherical equation ρ=2acosϕ, wherea>0. As with the graph of the cone with cylindrical equationz=2r, note that the equation is independent ofθ. Thus, all sections of this surface made by slicing with the half-planeθ =cmust be the same. If we compile values as in the adjacent table, then the section of the surface in the half-planeθ =0 is as shown in Figure 1.103. Since this section must be identical in all other constant-θ half-planes, we see that this surface appears to be a sphere of radiusatangent to thex y-plane, which is shown in Figure 1.104.
θ= 0
Section of = 2a cos ϕ ρ
(2a, 0, 0)
(a, /3, 0)π
Figure 1.103 The cross section ofρ=2acosϕin the half-plane θ=0.
θ= 0
θ= /4π θ= /3π θ= /2π
Figure 1.104 The graph ofρ= 2acosϕ.
The Cartesian equation of the surface is determined by multiplying both sides of the spherical equation byρand using the conversion equations in (7):
ρ=2acosϕ =⇒ ρ2 =2aρcosϕ
⇐⇒x2+y2+z2 =2az
⇐⇒x2+y2+(z−a)2=a2
by completing the square inz. This last equation can be recognized as that of a sphere of radiusawith center at (0,0,a) in Cartesian coordinates. ◆ EXAMPLE 10 NASA launches a 10-ft-diameter space probe. Unfortunately, a meteor storm pushes the probe off course, and it is partially embedded in the surface of Venus, to a depth of one quarter of its diameter. To attempt to reprogram the probe’s on-board computer to remove it from Venus, it is necessary to describe the embedded portion of the probe in spherical coordinates. Let us find the description desired, assuming that the surface of Venus is essentially flat in relation to the probe and that the origin of our coordinate system is at the center of the probe.
10/4'
10'
Surface of Venus Probe
Figure 1.105 The space probe of Example 10.
z
y
5/2 α 5
Figure 1.106 A slice of the probe of Example 10.
The situation is illustrated in Figure 1.105. The buried part of the probe clearly has symmetry about thez-axis. That is, any slice by the half-planeθ =constant looks the same as any other. Thus,θcan vary between 0 and 2π. A typical slice of the probe is shown in Figure 1.106. Elementary trigonometry indicates that for the angleαin Figure 1.106,
cosα=
5 2
5 = 1 2. z
y z = −5/2
Figure 1.107 Coordinate view of the cross section of the probe of Example 10.
Hence,α=cos−1 12 =π/3. Thus, the spherical angleϕ(which opens from the positivez-axis) varies fromπ−π/3=2π/3 toπ as it generates the buried part of the probe. Finally, note that for a given value of ϕbetween 2π/3 andπ,ρ is bounded by the surface of Venus (the planez= −52 in Cartesian coordinates) and the spherical surface of the probe (whose equation in spherical coordinates is ρ=5). See Figure 1.107. From the formulas in (7) the equationz= −52 corre- sponds to the spherical equationρ cosϕ= −52 or, equivalently, toρ= −52secϕ. Therefore, the embedded part of the probe may be defined by the set
(ρ, ϕ, θ) −5
2secϕ≤ρ≤5,2π
3 ≤ϕ≤π, 0≤θ <2π . ◆
Standard Bases for Cylindrical and Spherical Coordinates
In Cartesian coordinates, there are three special unit vectorsi,j, andkthat point in the directions of increasing x-, y-, and z-coordinate, respectively. We find corresponding sets of vectors for cylindrical and spherical coordinates. That is, in each set of coordinates, we seek mutually orthogonal unit vectors that point in the directions of increasing coordinate values.
In cylindrical coordinates, the situation is as shown in Figure 1.108. The vectorser,eθ, andez, which form thestandard basisfor cylindrical coordinates, are unit vectors that each point in the direction in which only the coordinate indicated by the subscript increases. There is an important difference between the standard basis vectors in Cartesian and cylindrical coordinates. In the former case,i,j, andkdo not vary from point to point. However, the vectorser andeθ dochange as we move from point to point.
ez
e
er
xi + yj
θ
Figure 1.108 The standard basis vectors for the cylindrical coordinate system.
Now we give expressions forer,eθ, andez. Since the cylindricalz-coordinate is the same as the Cartesian z-coordinate, we must have ez =k. The vector er must point radially outward from the z-axis with no k-component. At a point (x,y,z)∈R3 (Cartesian coordinates), the vector xi+yj has this prop- erty. Normalizing it to obtain a unit vector (see Proposition 3.4 of §1.3), we obtain
er = xi+yj x2+y2.
Wither andez in hand, it’s now a simple matter to defineeθ, since it must be perpendicular to bother andez. We take
eθ =ez×er = −yi+xj x2+y2.
(The reason for this choice of cross product, as opposed toer×ez, is so thateθ points in the direction ofincreasingθ.) To summarize, and using the cylindrical to Cartesian conversions given in (3),
er = xi+yj
x2+y2 =cosθi+sinθj;
eθ = −yi+xj
x2+y2 = −sinθi+cosθj; (8) ez =k.
In spherical coordinates, the situation is shown in Figure 1.109. In particular, there are three unit vectorseρ,eϕ, andeθthat form thestandard basisfor spherical coordinates. These vectors all change direction as we move from point to point.
xi + yj + zk eθ eρ
eϕ
Figure 1.109 The standard basis vectors for the spherical
coordinate system.
We give expressions foreρ,eϕ, andeθ. Since theθ-coordinates in both spher- ical and cylindrical coordinates mean the same thing,eθin spherical coordinates is given by the value ofeθ in (8). At a point (x,y,z), the vectoreρ should point
from the origin directly to (x,y,z). Thus,eρ may be obtained by normalizing xi+yj+zk. Finally,eϕ is nothing more thaneθ×eρ. If we explicitly perform the calculations just described and make use of the conversion formulas in (7), the following are obtained:
eρ= xi+yj+zk
x2+y2+z2 =sinϕ cosθi+sinϕ sinθj+cosϕk;
eϕ = x zi+yzj−(x2+y2)k x2+y2
x2+y2+z2
=cosϕ cosθi+cosϕ sinθj−sinϕk; (9)
eθ = −yi+xj
x2+y2 = −sinθi+cosθj.
Although the results of (8) and (9) will not be used frequently, they will prove helpful on occasion.
Hyperspherical Coordinates (optional)
There is a way to provide a set of coordinates forRn that generalizes spherical coordinates onR3. Forn≥3, thehyperspherical coordinatesof a pointP ∈Rn are (ρ, ϕ1, ϕ2, . . . , ϕn−1) and are defined by their relations with the Cartesian coordinates (x1,x2, . . . ,xn) of Pas
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
x1=ρsinϕ1sinϕ2· · ·sinϕn−2cosϕn−1
x2=ρsinϕ1sinϕ2· · ·sinϕn−2sinϕn−1
x3=ρsinϕ1sinϕ2· · ·sinϕn−3cosϕn−2
x4=ρsinϕ1sinϕ2· · ·sinϕn−4cosϕn−3
...
xn =ρcosϕ1
. (10)
To be more explicit, in equation (10) above we take
xk =ρsinϕ1 sinϕ2· · ·sinϕn−k cosϕn−k+1 fork=3, . . . ,n.
Note that whenn=3, the relations in (10) become
⎧⎪
⎨
⎪⎩
x1 =ρsinϕ1cosϕ2
x2 =ρsinϕ1sinϕ2
x3 =ρcosϕ1
.
These relations are the same as those given in (7), so hyperspherical coordinates are indeed the same as spherical coordinates whenn=3.
In analogy with (5), it is standard practice to impose the following restrictions on the range of values for the coordinates:
ρ≥0, 0≤ϕk ≤π fork=1, . . . ,n−2, 0≤ϕn−1 <2π. (11)
Then, with these restrictions, we can convert from hyperspherical coordinates to Cartesian coordinates by means of the following formulas:
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
ρ2=x12+x22+ · · · +xn2 tanϕ1=!
x12+ · · · +x2n−1/xn tanϕ2=!
x12+ · · · +x2n−2/xn−1 ...
tanϕn−2 =!
x12+x22/x3 tanϕn−1 =x2/x1
. (12)
Hyperspherical coordinates get their name from the fact that the (n−1)- dimensionalhypersurfaceinRn defined by the equationρ=ρ0, whereρ0 is a positive constant, consists of points on thehypersphereof radiusρ0 defined in Cartesian coordinates by the equation
x12+x22+ · · · +xn2=ρ02.