Solved problems on line integrals

This article provides solutions to problems on line integrals of vector fields. It covers topics like calculating line integrals, finding potential functions, and determining if a vector field is conservative. The problems are solved step-by-step, making it easy for students to understand the concepts. The article is relevant for students studying calculus, vector calculus, and engineering mathematics.

Q1)
F(x, y) = (exp(y), xexp(y) – 2ysin(y2)
r(t) = (t3, cos(πt)) where -1 ≤ t ≤1
x = t3, y = cos(πt)
r’(t) = 3t2, -πsin(πt)
F(r(t)) = (exp(y), xexp(y) – 2ysin(y2) * (3t2, -πsin(πt))
= ∫
C
❑
F ( x , y )∗dr = ∫
−1
1
¿ ¿exp(y), xexp(y) – 2ysin(y2) * (3t2, -πsin(πt))dt
= ∫
−1
1
3 t2 ¿ ¿
= ∫
−1
1
3 t2 exp( cos ( πt ) )−π sin ( πt)t3 exp (cos ( πt ) +2 π sin (πt )cos ( πt ) sin ( cos ( πt ) )2
] dt
= [ cos ( 2 πt ) +1
2 +t3 exp ¿ ¿]-11
= 1.03678 – 0.6322
= 0.40458
Q2)
F(x, y, z) = (z-x, -y, z)
r(t) = (t, t, sin(t))
x = t, y = t, z = sin(t)
r’(t) = 1, 1, cos(t)
F(r(t)) = sin(t)-t, -t, sin(t)
= ∫
γ
❑
F ( x , y , z )∗dr = ∫
0
1
¿ ¿sin(t)-t, -t, sin(t)]*[1, 1, cos(t)]dt
= ∫
0
1
¿ ¿ 1(sin(t)-t)+1* -t +cos(t)* sin(t)]dt

= ∫
0
1
¿ ¿ (sin(t) – 2t +cos(t)sin(t)]dt
= [ sin2 ( t ) −2(cos ( t ) +t2 )
2 +C]01
= 0.000304586 – 2*1.99985
= -1.9997
≈−2
Q3)
F(x, y)= [cos)x)-y(x2), xy2,ln(y+1)]
r(t) = (cos(t), sin(t))
X = cos(t), y = sin(t)
r’(t) = [-sin(t), cos(t)]
F(r(t)) = cos(cos(t)) – sin(t)(cos(t))2, cos(t)(sin(t))2 + ln(sin(t) + 1)
= ∫
γ
❑
F ( x , y )∗dr = ∫
0
1
¿ ¿ cos(cos(t)) – sin(t)(cos(t))2, cos(t)(sin(t))2 + ln(sin(t) + 1)]* [-sin(t),
cos(t)]dt
= ∫
0
1
¿ ¿
[(sin(t)+1)ln(sin(t)+1)+sin(cos(t))- sin(4t)/16 – sin(t)+t/4+C]0π
= -0.034905 + π
4
Q4)
a) U(x, y, z) = (xyz – x2, y2 + z2, 2xz)
∇ u= ( ∂
∂ x , ∂
∂ y , ∂
∂ z )∗(xyz – x 2, y 2+ z 2 , 2 xz )
∂
∂ x ( xyz−x2 ) + ∂
∂ y ( y2+z2 ) + ∂
∂ z (2 xz)
yz−2x+2y+2x
¿ yz +2 y
if r = ( x , y , z ) ∧r=¿ r ∨¿

Then the divergenceof F is 0
b) U(x, y, z) = ln(r)r
∇ u= ( ∂
∂ x , ∂
∂ y , ∂
∂ z )∗(ln ( x , y , z ) √x + y + z)
= ∂
∂ x ( lnx √x + y + z ) + ∂
∂ y ( lny √x + y + z ) + ∂
∂ z ( lnz √x + y +z )
= xlnx+2( x + y + z)
2 x √x + y + z + ylny +2( x + y +z)
2 y √x + y +z + zlnx+ 2( x+ y+ z )
2 z √x + y +z
= xyzlnx+2 yz ( x + y + z ) + xyzlny+ 2 xz ( x+ y + z ) + xyzlnz +2 zy ( x + y + z )
2 xyz √ x+ y+ z
if r = ( x , y , z ) ∧r=¿ r ∨¿
Then the divergenceof F is 1
¿ r ∨¿ ¿
Q5)
U= (y2z2 + xz, x2 +y, z)
X2 + y2= 4z2
Z = √ x2 + y2
4
Flux F = u
div(F) = ( ∂
∂ x , ∂
∂ y , ∂
∂ z )∗( y2 z2 + xz , x2+ y , z )
∂
∂ x ( y2 z2 + xz ) + ∂
∂ y ( x2 + y ) + ∂
∂ z (z )
= z + 2
Flux of F across Y = ∫∫ F∗ds(¿ Y )
= ∫
0
1
( z+ 2) dz
= 5/2
X = rcos θ , y=rsinθ

If z = 1
X2 + y2 = 4
r = 2
¿∫
0
2 π
∫
0
2
5
2 rdz dθ
5/2[ r2
2 ]02
= 5
∫
0
2 π
[5]dθ
= 5∗2 π
= 10 π
Q6)
Let take F = (y+z, x + z, x + y)
Curl of ⃗ F
| i j k
∂
∂ x
∂
∂ y
∂
∂ z
y + z x + z x + y
|
= i ( 1−1 ) − j ( 1−1 ) + k ( 1−1 ) =0
F is conservative and F = ( ∂ f
∂ x , ∂ f
∂ y , ∂ f
∂ z )
∂ f
∂ x = y +z=¿ f =∫ ( y+ z ) dx = (y + z)x + g(y, z)
f = (y + z)x + g(y, z)
fy = x + gy(y, z) = x + z
 Gy(y, z) = z => g(y, z) = ∫ zdy
 G(y, z) = zy + l(z)

f= x(y+z) + zy + l(z)
fz = x = y + lz(z) = x + y
lz(z) = 0 => l(z) = C
The potential function
f(x, y, z) = x(y+z) + yz + C