PHY2602 ASSIGNMENT 1 2023

PHY2602
ASSIGNMENT 1
2023

, QUESTION 1




Solution:


A = (5,1, −2) − (0, −1,3)

A = (5,2, −5)



Let û be the unit vector of A



A
û =
‖A‖



‖A‖ = √(5)2 + (2)2 + (−5)2

‖A‖ = √54

‖A‖ = 3√6



A
û =
‖A‖
(5,2, −5)
û =
3√6
5 2 5
û = ( , ,− ) unit vector of A
3√6 3√6 3√6

, QUESTION 2




Solution:


Volume of a cylinder by cylindrical coordinates:


.

Volume of a cylinder = ∭ r dr dθ dz
E




where E = {(r, θ, z): 0 ≤ z ≤ L, 0 ≤ θ ≤ 2π, 0 ≤ r ≤ R }


L 2π R

Volume of a cylinder = ∫ ∫ ∫ r dz dθ dz
0 0 0
L 2π R
r2
Volume of a cylinder = ∫ ∫ [ ] dθ dz
2 0
0 0
L 2π
R2 02
Volume of a cylinder = ∫ ∫ [ − ] dθ dz
2 2
0 0
L 2π
R2
Volume of a cylinder = ∫ ∫ dr dθ dz
2
0 0
L 2π
R2
Volume of a cylinder = ∫ ∫ 1 dθ dz
2
0 0
L
R2
Volume of a cylinder = ∫[θ]2π
0 dz
2
0
L
R2
Volume of a cylinder = ∫[2π − θ] dz
2
0
L
2πR2
Volume of a cylinder = ∫ 1 dz
2
0

Volume of a cylinder = πR2 [z]L0