Calculus II Solved Problems and Solutions | Desklib

Running head: ASSIGNMENT 1 1
Calculus II
Name
Institution

ASSIGNMENT 1 2
Question 1A
F= ( 5 x−2 y ) i+(4 x + y ) j
The equation of the circle with the center as origin is:
x2+ y2=32=9
x=rcos t, y=rsint
But r =3
x=3 cos t , y=3 sin t 0 ≤ t ≤ π
2
r(t) = {3 cos t ,3 sin t}
r’(t) = {−3 sin t , 3 cos t}
F= ( 5 x−2 y ) i+(4 x + y ) j
F (r (t))=¿ {15 cos t−6 sint , 12cos t+3 sin t}
F ( r ( t ) )−r ( t )=−3 sin t (15 cos t−6 sint )+3 cos t(12 cos t +3 sin t)
¿−45 sin t cos t+18 sin 2 t+36 cos 2t +9 sin t cos t
¿ 18+18 cos 2 t−36 sin t cos t
= 18 + 9(1 +cos2t) – 18sin2t
F(r(t)) (r’(t) = 27 + 9cot2t – 18sin2t
Work done = ∫ F ( r ( t ) ) . r ( t ) dt = ∫
0
π
2
(27+ 9cos 2t−18 sin 2 t)dt

ASSIGNMENT 1 3
¿ [27 t + 9
2 sin ( 2 t ) + 9cos (2 t)]
π
2
0
¿ 26 π
2 – 18=13 π −18=22.84
Therefore, work done¿ 22.84 Joules
Question 2A
V =2 y3 i+ ( 5+3 y2 ) j
Green theorem: the counterclockwise circulation of the field F=Mi+ Nj around a simple closed
curve C in a plane equals the double integral of the curve F over the region R enclosed by C.
That is,
∮
c
❑
F . dx = ∮
c
❑
Mdx+ Ndy = ∫∫
R
❑
( ∂ N
∂ x − ∂ M
∂ y )dA
V = 2y3i + (5 + 3xy2) j
We have M = 2y3, N = 5 + 3xy2
∂ N
∂ x =3 y2 ∂ M
∂ y =6 y2
= ∫∫
R
❑
( ∂ N
∂ x − ∂ M
∂ y )dA
¿∫∫¿ ¿)dx dy
= ∫∫−¿ 3 y2 dxdy ¿
Line joining O and B is y=x and that joining A and B is (1,0)∧(2,2) that is, x0 y0∧x1 y1 . Where