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Section 2.6 Cylindrical and Spherical
Coordinates
A) Review on the Polar Coordinates
The polar coordinate system consists of the origin O;the rotating ray or half line from O with unit tick. A point P in the plane can be uniquely described by its distance to the origin r = dist (P; O) and the angle μ; 0 · μ < 2 º :
θ
r
P(x,y)
O X
Y
We call (r; μ) the polar coordinate of P: Suppose that P has Cartesian (stan- dard rectangular) coordinate (x; y) :Then the relation between two coordinate systems is displayed through the following conversion formula:
Polar Coord. to Cartesian Coord.:
x = r cos μ y = r sin μ
Cartesian Coord. to Polar Coord.:
r =
p x^2 + y^2 tan μ =
y x 0 · μ < º if y > 0 ; 2 º · μ < º if y · 0 :
Note that function tan μ has period º; and the principal value for inverse tangent function is
¡
< arctan
y x
So the angle should be determined by
μ =
arctan
y x
; if x > 0
arctan
y x
; if x = 0; y > 0
¡
; if x = 0; y < 0
³^ Example 6.1.^ Find (a) Cartesian Coord. of^ P^ whose Polar Coord. is 2 ;
; and (b) Polar Coord. of Q whose Cartesian Coord. is (¡ 1 ; ¡1) : Sol. (a)
x = 2 cos
y = 2 sin
p 3 :
(b)
r =
p 1 + 1 =
p 2
tan μ =
= 1 =) μ =
or μ =
Since (¡ 1 ; ¡1) is in the third quadrant, we choose μ =
so
μp 2 ;
is Polar Coord.
Under Polar Coordinate system, the graph of any equation of two vari- ables r and μ is a curve. In particular, there are two families of coordinate curves that form Polar grid:
r = constant — circle centered at O with radius r μ = constant — ray with angle μ:
Thus, we readily have the conversion formula: x = r cos μ y = r sin μ z = z:
The reserve formula from Cartesian coordinates to cylindrical coordinates follows from the conversion formula from 2 D Cartesian to 2 D polar coordi- nates:
r^2 = x^2 + y^2 μ = arctan
y x
or arctan
y x
Example 6.2. (a) Plot the point with cylindrical coordinates (2; 2 º= 3 ; 1) ; and …nd its rectangular coordinates. (b) Find cylindrical coordinates of the point with rectangular coordinates (3; ¡ 3 ; ¡7) : Sol. (a) We …rst plot the point with 2 D polar coordinate (2; 2 º=3) in xy ¡ plane : r = 2; μ = 2º =3 = 120o: Then we raise it up vertically 1 unit. To …nd its rectangular coordinates, we use the formula
x = r cos μ = 2 cos
μ ¡
y = r sin μ = 2 sin
à (^) p 3 2
p 3
P(-1, √3 ,1)
Q(-1, √3 ,0)
θ = 2π/ O
Y
Z
X
(b)
r =
p x^2 + y^2 =
p 9 + 9 =
p 18
tan μ =
y x
= ¡ 1 ; arctan (¡1) = ¡
There are two choices for μ : ¡º = 4 or 3 º = 4 : Since (3; ¡3) is in the 4th quad- rant, we …nd that the cylindrical coordinates for the point with rectangular coord (3; ¡ 3 ; ¡7) is (^) ³ p 18 ; ¡º = 4 ; ¡ 7
As with the polar coordinate system, one …nds it very convenient and simple to represent many surface using cylindrical coordinates instead of the rectangular coordinate system. For instance, the coordinate planes, under the cylindrical coordinate system, consists of three families surfaces described as follows:
² Equation r = constant represents a circular cylinder with z ¡ axis as its symmetric axis.
r = 1
² Equation μ = constant represents a half-plane originating from z ¡axis with the constant angle to zx ¡ plane y = x
Therefore, equations for cylinder-like surfaces may be much easier using the cylindrical coordinate system. Example 6.3. (a) Describe he surface whose cylindrical equation is z = r: (b) Find the cylindrical equation for the ellipsoid 4 x^2 + 4y^2 + z^2 = 1. (c) Find the cylindrical equation for the ellipsoid x^2 + 4y^2 + z^2 = 1: Solution: (a)
z = r =) z^2 = r^2 =) z^2 = x^2 + y^2
This a cone with its axis on z ¡ axis: (b)
4 x^2 + 4y^2 + z^2 = 1 =) 4 r^2 + z^2 = 1
(c) If we use the cylindrical coordinate as we introduced above, we would get
x^2 + 4y^2 + z^2 = 1 r^2 cos^2 μ + 4r^2 sin^2 μ + z^2 = 1
or r^2 + 3r^2 sin^2 μ + z^2 = 1:
However, if we change the axis of the cylindrical coordinate system to y¡axis; i.e.,
z = r cos ' x = r sin ' y = y;
where (r; ') is the polar coordinate of the projection onto zx ¡ plane; then the equation between
x^2 + 4y^2 + z^2 = 1 =) 4 y^2 + r^2 = 1:
C) Spherical Coordinate System.
In the spherical coordinate system, a point P (x; y; z), whose Cartesian coordinates are (x; y; z) ; is described by an ordered triple (Ω; μ; ¡) ; where
Ω> 0 ; 0 · μ · 2 º ; 0 · ¡ · º
are de…ned as follows.
² Ω= dist (P; O)
² μ is de…ned in the same way in the cylindrical coordinate system: Angle from zx ¡ plane; counter-clockwise, to the half-plane originating from z ¡ axis and containing P
² ¡ = angle from positive z ¡ axis to vector
OP :
φ
ρ
θ (^) r
P(x,y,z)
O Y
Z
X Q(x,y,0)
z
Note that when P is on z ¡ axis; ¡ = 0; and
¡ increases from 0 to
as P moves closer to xy ¡ plane;
and ¡ keeps increasing as P moves below xy ¡ plane; and ¡ reaches the maximum value º when P is on the negative z ¡ axis:
Conversion formula (rectangular $ cylindrical $ spherical)
r = 2
ρ = 2 φ = π/
Q
θ = π/
P(x,y,z)
O Y
Z
X
x = Ωsin ¡ cos μ = 2 sin
cos
p 6 2
y = Ωsin ¡ sin μ = 2 sin
sin
p 6 2 z = Ωcos ¡ = 2 cos
(b)
p x^2 + y^2 + z^2 =
r³ 2
p 3
p 16 = 4
μ =
; since
p 3 ; ¡ 2
is on positive y ¡ axis
cos ¡ =
z Ω
Ans: Spherical coord. =
μ 4 ;
The coordinate surfaces are:
² Ω= Ω 0 (constant) is a sphere centered at the origin with radius Ω 0
² μ = μ 0 (constant) is a half-plane originated from z ¡ axis with angle μ 0 to zx ¡ plane (the same as in the cylindrical coordinate system
² ¡ = ¡ 0 (constant) is a circular cone with z ¡ axis as its symmetric axis and the opening angle ¡ 0
In fact, if we convert ¡ = ¡ 0 into the rectangular coordinate system, we have
x = Ωsin ¡ 0 cos μ y = Ωsin ¡ 0 sin μ z = Ωcos ¡ 0 :
We can eliminate μ by
x^2 + y^2 = Ω^2 sin^2 ¡ 0 z^2 = Ω^2 cos^2 ¡ 0 :
We then eliminate Ωby dividing the …rst equation by the second
x^2 + y^2 z^2
= tan^2 ¡ 0 or
x^2 tan^2 ¡ 0
y^2 tan^2 ¡ 0
z^2 1
which is a cone. Again, these three families of coordinate surfaces form the spherical grids:
Spherical Grids
(c) ¡ =
(d) μ =
- Convert into (i) cylindrical equation and (ii) spherical equation. You must simply your solutions.
(a) z = 2x^2 + 2y^2 (b) x^2 + y^2 ¡ 2 z^2 = 3 (c) x^2 + 2y^2 ¡ z^2 = 3
- (optional) Identify the surface by converting into rectangular equation.
(a) r = 3 cos μ (b) Ω= 3 cos ¡ (c) r^2 ¡ 2 z^2 = 1:
