Calculus and Analysis Homework: Vector Calculus & Fourier Series

This document offers a comprehensive solution to a Calculus assignment, addressing key concepts in vector calculus and Fourier series. The solution begins with determining if a vector field is conservative, calculating line integrals, and applying Stokes' theorem. It then proceeds to use the Gauss divergence theorem to evaluate a surface integral. The assignment also includes a detailed analysis of Fourier series, including finding the Fourier series representation of a given function and calculating the coefficients for a complex exponential series. The solutions are presented step-by-step, making them suitable for students seeking to understand and master these fundamental concepts in calculus and analysis.

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