Solutions to Vector Calculus Problems

This JSON response provides solutions to vector calculus problems including finding scalar potential function, using Stokes theorem, Gauss divergence theorem, and Fourier series. The solutions are explained step-by-step with mathematical equations and integrals.

Solution
Q1)
F(x, y, z) = (2xz + zsiny, xzcosy, x2 + xsiny)
Curl F =
| ^i ^j ^k
∂
∂ x
∂
∂ y
∂
∂ z
∂ xz + zsiny xzcosy x2+xsiny |
= (xcosy – xcosy) ^i−¿
= 0 ^i−0 ^j+0 ^k
= 0
F is conservative in R3
To determine a scalar potential function
∂ f
∂ y =¿2xz + zsiny,
∂ f
∂ y =xzcosy ,
∂ f
∂ y =x2 +xsiny
Integrating ∂ f
∂ y =¿2xz + zsiny
f = x2z = xysiny + g(y,z)
The partial derivative of f = x2z = xysiny + g(y,z) with respect to y
∂ f
∂ y =¿xycosy + ∂ g ( y , z )
∂ y
Comparing to ∂ f
∂ y =xzcosy
xzcosy + ∂ y
∂ y = xzcosy
∂ y
∂ y = 0

g(y, z) = h(z)
Partial derivative of f = x2z = xysiny + g(y,z) with respect to z
∂ f
∂ z =x2 + xsiny+h' (z)
Comparing this to ∂ f
∂ y =x2 + xsiny
x2+ xsiny+h' ( z )=x2+ xsiny
= h’(z) = 0
= h(z) = C where C is a constant.
f(x, y, z) = x2z + xzsiny + C
By fundamental theorem of line integral
∫
C
❑
F . dr=f (−1 , 0 , 4 )−f (1 , 0 ,0)
f(-1, 0, 4) = 4 + C
f(1, 0, 0) = 0 + C
then f(-1, 0, 4) – f(1, 0, 0) = 4
therefore,
∫
C
❑
F . dr=4
Q2)
Using stokes theorem
∬
S
❑
( ∇∗F )∗nds=∫
C
❑
F .dr
The parabolic z = 4 – x2 – y2 intersects xy- plane as a circle x2 + y2 = 4
r(t) = (2cost, 2sint, 0), 0 > t ≤ 2 π

r’(t) = (-2sint, 2cost, 0)
∫
C
❑
F . dr=∫
t
❑
F ( r ( t ) )∗r ' ( t ) dt
= ∫
t
❑
(2 sin t ¿ e0−2 sin t¿, 2 cos t∗e0 ,1+4 sin t cos t e0)(−2 sin t ,2 cos t ,0)∗dt ¿
= ∫
t
❑
(0,2 cos t¿ ,1+4 sin t cos t) (−2 sin t , 2 cos t ,0 ) dt ¿
= ∫
t
❑
4 cos2 t dt=4∫
t
❑ 1+cos 2 t
2 dt
= 2[(t)02π + ( sin 2t
2 )02π]
= 2[2 π + 0 ¿
= 4 π
= ∬
S
❑
( ∇∗F )∗nds=4 π
Q3)
By Gauss divergence theorem
∬
S
❑
F . ^n ds=∭
V
❑
¿⃗ F dv, where V is the volume bounded by sphere S
divF = ∇∗F= ∂
∂ x ( x ) + ∂
∂ y ( 2 y ) + ∂
∂ z ( x z2 )
= 1 + 2 + 2xz
= 3 + 2xz
∬
S
❑
F . ^n ds=∭
V
❑
( 3+2 x z ) dxdydz …………….eq1
Converting this to spherical polar coordinate system, then

x – 2 = rsin θcos ∅ , y = rsinθcos ∅ , z = rcosθ, 0 ≤ r ≤ 1 , 0≤ θ ≤ π , 0 ≤ ∅ <2 π
3 + 2xz = 3 + 2(2 + rsin θcos ∅ ) rcoθ
=3 = 4rco θ + 2rsinθcosθcos∅
|J| = Jacobin = r2sinθ
∬
S
❑
F . ^n ds=∭
V
❑
( 3+2 xz ) dxdydz
= ∫
0
1
∫
0
π
∫
0
2 π
( 3+ 4 rcosθ+2 rsinθcos ∅ cosθ ) ∨J ∨d ∅ dθdr
= ∫
0
1
∫
0
π
∫
0
2 π
( 3+ 4 rcosθ+2 rsinθcos ∅ cosθ ) r2 sinθd ∅ dθdr
= ∫
0
1
∫
0
π
∫
0
2 π
( 3 r2 sinθ +4 r3 sinθcosθ +2 r3 sin2 θ cos ∅ cosθ ) d ∅ dθdr
= ∫
0
1
∫
0
π
¿ ¿ ¿02π + (4 r3 sinθcosθ )(∅ )02π + 2r3sin2
θcos θ(sin ∅ ¿02π)dθdr
Since sin2π= 0 = sin0
= ∫
0
1
∫
0
π
2 π ( 3 r2 sinθ+ 4 r3 sinθcosθ ) dθdr
= 2π∫
0
1
∫
0
π
( 3 r 2 sinθ+ 42 sin2 θ ) dθdr
Therefore, sin2 θ=2 sinθcosθ
= 2π∫
0
1
¿ ¿0π + 2r3(−cos 2 θ
2 ¿0π]dr
Since cosπ = -1, cos0 = 1, cos2π = 1
= 2π∫
0
1
( −3 r2 ( cosπ−cos 0 ) −r3 ( cos 2 π−cos 0 ) ) dr

= 2π∫
0
1
−3 r 2 ( −1−1 ) −r3 (1−1)¿ dr ¿
= 2π∫
0
1
6 r2 dr
= 12 π∫
0
1
r2 dr
= 12 π [ r 3
3 ]10
= 12π[1/3 – 0]
= 4π
Q4)
a) We are given f(x) = x4 - 2π2 x2 ,−π ≤ x ≤ π
F(x) = x4 - 2 π2 x2
f(-x) = (-x)4-2π2(-x)2 = x4 -2π2x2
it is a exists even function
f(x) = a 0
2 +∑
n=1
∞
ancosnx
bn = 0
a0 = 2
π ∫
0
π
f ( x ) dx= 2
π ∫
0
π
( x4−2 π2 x2 ) dx
= 2
π ( x5
5 − 2 π 2 x3
3 )0π
= 2
π ( 3 π 5−10 π 5
15 )
= ( −14 π4
15 )

an = 2
π ∫
0
π
f ( x ) cosnxdx
= 2
π ∫
0
π
( x4−2 π2 x2 ) cosnxdx
= −2
π ¿– 24x cosnx
n4 + 24∗sinnx
n5 ¿0π
= 2
π [0+0−0−24 π (−1 )n
n4 +0−0]
= −48 (−1 )n
n4 = 48 (−1 )n+ 1
n4
Required fourier series
F(x) = a 0
2 +∑
n=1
∞
ancosnx
= −7 x4
15 +∑
n=1
∞ 48 (−1 )n+1
n4 cosnx
x4 - 2 π 2 x2=−7 x4
15 − 48
14 cosx − 48
24 cos 2 x + 48
34 cos 3 x− 48
44 cos 4 x +…
b) put x = π
x4 - 2 π 2 x2=−48( 1
14 + 1
24 + 1
34 + 1
44 …)
−15 π4 +7 π4
15 ∗−1
48 = 1
14 + 1
24 + 1
34 + 1
44 …
8 π4
15 ∗1
48 = 1
14 + 1
24 + 1
34 + 1
44 …

π4
90 = 1
14 + 1
24 + 1
34 + 1
44 …
Q5)
f(t) = ∑
n=−∞
∞
Cn*einω0 t
Cn = 1
π ∫
−π
π
f (t)einω0 t dt
Calculate Cn for the given function
Cn = 1
π ∫
−π
π
t e−2 t e−inω 0 t dt
Integrate by parts ∫ fg '=fg−∫f ' g
f(t) = t:g’(t) = e−inω0 t −2 t
The integral will be
Solving the indefinite integral
Cn = 1
π ¿
U = -in ω0t – 2t
dt = 1
−inω0−2 du
Cn = 1
π [ −t e−inω0 t −2t
inω 0+2 − e−inω0 t −2 t
(−inω 0+2)(inω 0+2) ]-ππ
Cn = 1
π (−( π e4 π nω 0 – 2i π e4 π +i e4 π )e2 iπnω 0 +πnω 0−2 iπ −i¿ (e−iπnω 0−2 π )¿¿ i n2 ω2 0+ 4 nω 0−4 i)
Cn= − ( ( π e4 π nω 0 – 2i π e4 π +i e4 π ) e2 iπnω 0+ πnω 0−2 iπ −i¿ (e−iπnω 0−2 π )¿¿ i n2 ω2 0 π +4 nω 0 π−4 i π )
From this coefficient we can be able to find the Fourier complex exponential series