MAT2615 Assignment 4 Solutions 2026
UNISA
At the end of this assignment there is all drawings, locate them and draw each at its
respective question. (drawings are numbered) thank you……
, Question 1
Given
𝐹(𝑥, 𝑦) = (6, 𝑥𝑦 − 12 3𝑥 2 )
and
𝑥 2 + 𝑦 2 = 4,
from (2, 0)to (−, 20)anticlockwise.
(a) Line integral by parametrisation
Parametrise the circle:
𝑟(𝑡) = (2, cos 𝑡 2 sin 𝑡), 0 ≤ 𝑡 ≤ 𝜋.
Therefore,
𝑥 = 2 cos 𝑡 , 𝑦 = 2 sin 𝑡
𝑑𝑥 = −2 sin 𝑡 𝑑𝑡, 𝑑𝑦 = 2 cos 𝑡 𝑑𝑡.
Now,
6𝑥𝑦 − 12 = 6(2 cos 𝑡)(2 sin 𝑡) − 12
= 24 cos 𝑡 sin 𝑡 − 12.
Also,
3𝑥 2 = 3(2 cos 𝑡)2 = 12cos 2 𝑡.
Hence
∫𝐹 ⋅ 𝑑𝑟 = ∫( 6𝑥𝑦 − 12) 𝑑𝑥 + 3𝑥 2 𝑑𝑦.
𝐶 𝐶
Substitute:
𝜋
= ∫ ( 24 cos 𝑡 sin 𝑡 − 12)(−2 sin 𝑡) + 12cos 2 𝑡(2 cos 𝑡) 𝑑𝑡
0
𝜋
= ∫ (−48 cos 𝑡 sin2 𝑡 + 24 sin 𝑡 + 24cos3 𝑡) 𝑑𝑡.
0
UNISA
At the end of this assignment there is all drawings, locate them and draw each at its
respective question. (drawings are numbered) thank you……
, Question 1
Given
𝐹(𝑥, 𝑦) = (6, 𝑥𝑦 − 12 3𝑥 2 )
and
𝑥 2 + 𝑦 2 = 4,
from (2, 0)to (−, 20)anticlockwise.
(a) Line integral by parametrisation
Parametrise the circle:
𝑟(𝑡) = (2, cos 𝑡 2 sin 𝑡), 0 ≤ 𝑡 ≤ 𝜋.
Therefore,
𝑥 = 2 cos 𝑡 , 𝑦 = 2 sin 𝑡
𝑑𝑥 = −2 sin 𝑡 𝑑𝑡, 𝑑𝑦 = 2 cos 𝑡 𝑑𝑡.
Now,
6𝑥𝑦 − 12 = 6(2 cos 𝑡)(2 sin 𝑡) − 12
= 24 cos 𝑡 sin 𝑡 − 12.
Also,
3𝑥 2 = 3(2 cos 𝑡)2 = 12cos 2 𝑡.
Hence
∫𝐹 ⋅ 𝑑𝑟 = ∫( 6𝑥𝑦 − 12) 𝑑𝑥 + 3𝑥 2 𝑑𝑦.
𝐶 𝐶
Substitute:
𝜋
= ∫ ( 24 cos 𝑡 sin 𝑡 − 12)(−2 sin 𝑡) + 12cos 2 𝑡(2 cos 𝑡) 𝑑𝑡
0
𝜋
= ∫ (−48 cos 𝑡 sin2 𝑡 + 24 sin 𝑡 + 24cos3 𝑡) 𝑑𝑡.
0