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Section 2.6 Cylindrical and Spherical Coordinates

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Section 2.6 Cylindrical and Spherical Coordinates 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 < 2¼ : · Y P(x,y) r θ O X 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: x = r cos µ Polar Coord. to Cartesian Coord.: y = r sin µ ½ r = x2 + y2 Cartesian Coord. to Polar Coord.: y tan µ = ( p x 0 µ < ¼ if y > 0, 2¼ µ < ¼ if y 0. · · · Note that function tan µ has period ¼, and the principal value for inverse tangent function is ¼ y ¼ < arctan < . ¡ 2 x 2 1 So the angle should be determined by y arctan , if x > 0 xy 8 arctan + ¼, if x < 0 µ = > ¼ x > > , if x = 0, y > 0 < 2 ¼ , if x = 0, y < 0 > ¡ 2 > > Example 6.1. Fin:>d (a) Cartesian Coord. of P whose Polar Coord. is ¼ 2, , and (b) Polar Coord. of Q whose Cartesian Coord. is ( 1, 1) . 3 ¡ ¡ ³ So´l. (a) ¼ x = 2 cos = 1, 3 ¼ y = 2 sin = p3. 3 (b) r = p1 + 1 = p2 1 ¼ ¼ 5¼ tan µ = ¡ = 1 = µ = or µ = + ¼ = . 1 ) 4 4 4 ¡ 5¼ Since ( 1, 1) is in the third quadrant, we choose µ = so ¡ ¡ 4 5¼ p2, is Polar Coord. 4 µ ¶ 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. 2 B) Cylindrical Coordinate System The cylindrical coordinate system basically is a combination of the polar coordinate system xy plane with an additional z coordinate vertically. In the cylindrical coo¡rdinate system, a point P (x,¡y, z) , whose Cartesian coordinate is (x, y, z) , is assigned by the ordered triple (r, µ, z) , where (r, µ) is the polar coordinate of (x, y) , the vertical projection along z axis of P onto xy plane. ¡ ¡ Z P(x,y,z) z O Y θ r X Q(x,y,0) 3 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 2D Cartesian to 2D polar coordi- nates: r2 = x2 + y2 y y µ = arctan or arctan + ¼. x x Example 6.2. (a) Plot the point with cylindrical coordinates (2, 2¼/3, 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 w¡ith ¡2D polar coordinate (2, 2¼/3) in xy plane : r = 2, µ = 2¼/3 = 120o. Then we raise it up vertically 1 unit. To ¡…nd its rectangular coordinates, we use the formula 2¼ 1 x = r cos µ = 2 cos = 2 = 1 3 ¡2 ¡ µ ¶ 2¼ p3 y = r sin µ = 2 sin = 2 = p3 3 Ã 2 ! P(-1, √3 ,1) Z Q(-1, √3 ,0) θ = 2π/3 O Y X 4 (b) r = x2 + y2 = p9 + 9 = p18 y 3 ¼ tan µ = p= ¡ = 1, arctan ( 1) = . x 3 ¡ ¡ ¡ 4 There are two choices for µ : ¼/4 or 3¼/4. Since (3, 3) is in the 4th quad- rant, we …nd that the cylindr¡ical coordinates for the ¡point with rectangular coord (3, 3, 7) is ¡ ¡ p18, ¼/4, 7 . ¡ ¡ As with the polar coordi³nate system, on´e …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: Equation r = constant represents a circular cylinder with z axis as ² its symmetric axis. ¡ r = 1 Equation µ = constant represents a half-plane originating from z axis ² with the constant angle to zx plane y = x ¡ ¡ 5 ¼ µ = 0, µ = 4 z = constant represents a plane parallel to xy plane, the same as in ² the rectangular system. ¡ Put all three families together, we have the cylindrical grids: Cylindrical Grids 6 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 4x2 + 4y2 + z2 = 1. (c) Find the cylindrical equation for the ellipsoid x2 + 4y2 + z2 = 1. Solution: (a) z = r = z2 = r2 = ) ) z2 = x2 + y2 This a cone with its axis on z axis. (b) ¡ 4x2 + 4y2 + z2 = 1 = ) 4r2 + z2 = 1 (c) If we use the cylindrical coordinate as we introduced above, we would get x2 + 4y2 + z2 = 1 r2 cos2 µ + 4r2 sin2 µ + z2 = 1 or r2 + 3r2 sin2 µ + z2 = 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 ¡ x2 + 4y2 + z2 = 1 = ) 4y2 + r2 = 1. C) Spherical Coordinate System. 7 In the spherical coordinate system, a point P (x, y, z), whose Cartesian coordinates are (x, y, z) , is described by an ordered triple (½, µ, Á) , where ½> 0, 0 µ 2¼, 0 Á ¼ · · · · are de…ned as follows. ½= dist (P, O) ² µ is de…ned in the same way in the cylindrical coordinate system: Angle ² from zx plane, counter-clockwise, to the half-plane originating from z axis¡and containing P ¡ Á = angle from positive z axis to vector ¡O!P . ² ¡ Z P(x,y,z) ρ z φ O Y θ r X Q(x,y,0) Note that when P is on z axis, Á = 0, and ¡ ¼ Á increases from 0 to as P moves closer to xy plane, 2 ¡ and Á keeps increasing as P moves below xy plane, ¡ and Á reaches the maximum value ¼ when P is on the negative z axis. ¡ Conversion formula (rectangular cylindrical spherical) $ $ 8 x = r cos µ = ½sin Á cos µ y = r sin µ = ½sin Á sin µ z = r cot Á = ½cos Á ½= x2 + y2 + z2 y tan µ = p x z cos Á = . x2 + y2 + z2 Note that when we determine µ, wepalso need to consider which quadrant the point is in. More precisely, y arctan , if x > 0 xy 8 arctan + ¼, if x < 0 µ = > ¼ x . > > , if x = 0, y > 0 < 2 ¼ , if x = 0, y < 0 > ¡ 2 > However, Á can be det>ermined uniquely as :> z Á = arccos . Ã x2 + y2 + z2 ! Example 6.4. (a) Plot the pointpwith spherical coordinates (2, ¼/4, ¼/3) , and …nd its rectangular coordinates. (b) Find the spherical coordinates for the point with rectangular coordinates 0, 2p3, 2 . Sol: (a) We …rst plot the point Q on xy p¡lane with polar coordinate ¡ ¡ ¢ (2, ¼/4) . We then rotate ¡O!Q in the vertical direction (i.e., the rotation re- mains in the plane spanned by z-axis and line OQ) till its angle with z axiz reaches ¼/3. ¡ 9 Z P(x,y,z) ρ = 2 φ = π/3 O Y r = 2 θ = π/4 X Q ¼ ¼ p6 x = ½sin Á cos µ = 2 sin cos = 3 4 2 ³¼ ´ ³¼ ´ p6 y = ½sin Á sin µ = 2 sin sin = 3 4 2 ¼ ³ ´ ³ ´ z = ½cos Á = 2 cos = 1. 3 ³ ´ (b) 2 ½= x2 + y2 + z2 = 2p3 + 4 = p16 = 4 ¼ r³ ´ µ = p, since 0, 2p3, 2 is on positive y axis 2 ¡ ¡ z 2 ³ 1 ´ 2¼ cos Á = = ¡ = = Á = . ½ 4 ¡2 ) 3 ¼ 2¼ Ans: Spherical coord. = 4, , . 2 3 The coordinate surfacµes are: ¶ ½= ½0 (constant) is a sphere centered at the origin with radius ½0 ² µ = µ0 (constant) is a half-plane originated from z axis with angle ² µ to zx plane (the same as in the cylindrical coor¡dinate system 0 ¡ 10 Á = Á0 (constant) is a circular cone with z axis as its symmetric ² ¡ axis and the opening angle Á0 In fact, if we convert Á = Á0 into the rectangular coordinate system, we have x = ½sin Á0 cos µ y = ½sin Á0 sin µ z = ½cos Á0. We can eliminate µ by 2 2 2 2 x + y = ½ sin Á0 2 2 2 z = ½ cos Á0. We then eliminate ½by dividing the …rst equation by the second 2 2 2 2 2 x + y 2 x y z 2 = tan Á0 or 2 + 2 = z tan Á0 tan Á0 1 which is a cone. Again, these three families of coordinate surfaces form the spherical grids: Spherical Grids 11 Example 6.5. Find the spherical equation for the hyperboloid of two sheets x2 y2 z2 = 1. Solutio¡n: B¡y direct substitution, we obtain, under the standard spherical coordinate system (½sin Á cos µ)2 (½sin Á sin µ)2 (½cos Á)2 = 1 ¡ ¡ or ½2 sin2 Á cos2 µ sin2 Á sin2 µ cos2 Á = 1. ¡ ¡ Example 6.6. ¡Find a rectangular equation for the¢ surface whose spher- ical equation is ½= 2 sin Á sin µ. Solution: We want to eliminate the spherical variables ½, µ, Á and replace them with x, y, z. To this end, we multiply the equation by ½to obtain ½2 = 2½sin Á sin µ. From the conversion formula, we have x2 + y2 + z2 = 2y or x2 + (y 1)2 + z2 = 1.
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