Solved Examples/Numericals On Coordinate System & Transformation - 2.

Q.7 Calculate the distance between the following pair of points?

a) (2, 1, 5) and (6, -1, 2)
b) (3, π/2, -1) and (5, 3π/2, 5)
c) (10, π/4, 3π/4) and (5, π/6, 7π/4)

Ans. a) Distance between two points in Cartesian co-ordinate is given as:

d2 = (x2 – x1) 2 +(y2 – y1) 2 +(z2 – z1) 2
= (6 – 2) 2 + (-1 -1) 2 + (2 – 5) 2
= (29) 1/2    => d = 5.38


b) Distance between two points in cylindrical co-ordinate system is given as:

d2 = ρ22 + ρ12 - 2ρ1ρ2 cos(φ2 – φ1) + (z2 – z1) 2
= 52 + 32 – 2.5.3 cos(3π/2 - π/2) + (5 +1) 2
= (100) 1/2  =>d = 10.


c) Distance between two points in spherical co-ordinate system is given as:

d2 = r22 + r12 – 2r1r2 cosθ2 cosθ1 + 2r1 r2 sinθ2 sinθ1 cos(φ2 – φ1)
= 102 + 52 – 2.5.10 cosπ/4 cosπ/6 + 2.5.10 sinπ/4 sinπ/6 cos(7π/4 - 3π/4)
= (99.12) 1/2  => d = 9.96



Q.8 Find the distance between A (2, π /6, 0) and B = (1, π /2, 2) ?
Ans:

The points are given in Cylindrical Co-ordinate (ρ, φ, z). To find the distance between two points, the co-ordinates are to be in Cartesian (rectangular). The corresponding rectangular co-ordinates are (ρ Cosφ, ρ Sinφ, z).

Position vector of A:

rA = x ax + y ay + z az
= ρ Cosφ ax + ρ Sinφ ay + z az
= 2cos φ/6 ax + 2 Sinφ/6 ay
= 1.73 ax + ay

Similarly,

Position vector of B:

rB = ρ Cosφ ax + ρ Sinφ ay + z az
= 1 Cosφ/2 ax + 1 Sinφ/2 ay + 2az
= ay + 2az


Distance Vector rAB = rB – rA = - 1.73 ax + 2az

Hence the distance between the two points is given as :
| rAB | = 2.64



Q.9 Using the differential length dl, for the length of each of the following curves:
a) ρ = 3, π /4 < φ < π /2, z = constant.
Ans:

dl = ρ dφ

L = dl = 3∫π /4π /2 dφ = 3 ( π /2 – π /4) = 3π /4 = 2.356


b) r = 1, θ = 30o, 0o < φ < 60o
Ans:

dl = rsinθ dφ

L = ∫dl = rsinθ∫0π /3 dφ = (1) sin 30o [ π /3 – 0] = 0.5236


c) r = 4, φ = constant, 30o < θ < 90o
Ans:

dl = rdθ

L = ∫dl = r∫π /6π /2dθ = 4 [ π /2 – π /6] = 4π /3 = 4.189



Q.10 Calculate the areas of the following surfaces using the differential surface area ds:
a) ρ = 2, π/3 < φ < π/2, 0 < z < 5.
Ans:

ds = ρ dφ dz

S = ∫ds = ∫ ∫ ρ dφ dz = 2 ∫π/3π/2 dφ  ∫05 dz = 2 (5) [π/2 – π/3] = 10π/6 = 5.236


b) r = 10, π/4 < θ < 2π/3 , 0 < φ < 2π
Ans:

ds = r2sinθ dθdφ

S = ∫ds = ∫ ∫ r2sinθ dθdφ = 102π/42π/3sinθdθ  ∫0dφ = 100(2π) |(-cosθ)|π/42π/3 = 7.584



Q.11 Use the differential volume dv to determine the volumes of the following regions:
a) 0 < x < 1, 1 < y <2, -3 < z < 3 
Ans:

dv = dx dy dz 

V = ∫ ∫ ∫ dx dy dz = ∫01dx  ∫12dy  ∫-33dz = (1) (2 - 1)(3 +3) = 6


b) 2 < ρ < 5, π/3 < φ < π, - 1 < z < 4
dv = ρ dρ dφ dz

V = ∫ ∫ ∫ ρ dρ dφ dz = ∫25ρdρ ∫π/3π dφ ∫-14 dz = | ρ2 /2 |25 (π – π /3) (4 + 1) = 35 π = 110



ALSO READ:

- Introduction To Coordinate System.

- Cartesian Coordinate System / Rectangular Coordinate System (x, y, z).

- Differential Analysis Of Cartesian Coordinate System.

- Circular Cylindrical Coordinate System (ρ, φ, z).

- Differential Analysis Of Cylindrical Coordinate System.

- Spherical Coordinate System ( r, θ , φ).

- Differential Analysis Of Spherical Coordinate System.

- Numericals / Solved Examples - Page 1.

- Numericals / Solved Examples - Page 2.


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