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Solution
Assignment # 7
Q 1. Let V be the region bounded by the surface x 2 + y 2 = 4 and the planes
z = 0 and z = 3 in three dimensional space and let S denote the surface
→ ^ ^ ^ → ^
of V . If F = x3 i + y 3 j + z 3 k , use Divergence Theorem to find ∫ F • n dS
S
^
OM
where n denote the unit outer normal to S.
Solution:
Divergence Theorem or Gauss’s Theorem states that
C
S.
Let V be a region in three dimensional space which is bounded by a closed
surface S (In three dimensional space such region V is a volume of the solid
N^
whose surface is S). Also suppose that n denote the unit outer normal to S at any
IA
→
point (x, y, z). If F is a vector function that has continuous partial derivatives on
→
⎛ →
⎞
⎜ ∇ • F ⎟ over V and the
AL
V then the volume integral of the divergence of F i-e
⎝ ⎠
→
surface integral of F over the surface S of V are related by
U
RT
→ ^ →
∫ F • n dS = ∫ ∇ • F dV
S V
Or
VI
→ ^ →
∫ F • n dS = ∫ div F dV
S V
Now come up to the solution of this question.
According to the given statement, V is a
cylinder of radius 2 and height 3 (as shown in
figure)
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→ ^ ^ ^
F = x3 i + y 3 j + z 3 k
→ ⎛ ∂ ^ ∂ ^ ∂ ^⎞ ⎛ 3^ 3 ^ 3 ^⎞
∇ • F = ⎜ i+ j+ k ⎟ • ⎜ x i+ y j+ z k ⎟
⎝ ∂x ∂y ∂z ⎠ ⎝ ⎠
∂x ∂y ∂z
3 3 3
= + +
∂x ∂y ∂z
= 3x 2 + 3 y 2 + 3z 2
= 3( x 2 + y 2 + z 2 )
So by Divergence Theorem
→ →
OM
^
∫ F • n dS = ∫ ∇ • F dV
S V
= 3∫ ( x 2 + y 2 + z 2 ) dV
C
V
S.
Use cylindrical co-ordinates to evaluate the volume integral at right hand side.
x = r cos θ
y = r sin θ N
z=z
IA
x 2 + y 2 + z 2 = r 2 cos 2 θ + r 2 sin 2 θ + z 2
= r2 + z2
dV = r dz dr dθ
U AL
Limits for
z is 0 to 3
RT
r is 0 to 2
θ is 0 to 2π
VI
To have better understanding of cylindrical co-ordinates, see the following link.
http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/bo
dy.htm
So, putting values
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→ ^ →
∫ F • n dS = ∫ ∇ • F dV = 3∫ ( x + y + z ) dV
2 2 2
S V V
2π 2 3
= 3∫ ∫ ∫ (r 2 cos 2 θ + r 2 sin 2 θ + z 2 ) r dz dr dθ
0 0 0
2π 2 3
= 3∫ ∫ ∫ (r (cos 2 θ + sin 2 θ ) + z 2 ) r dz dr dθ
2
0 0 0
2π 2 3
= 3∫ ∫ ∫ (r 3 + rz 2 ) dz dr dθ
0 0 0
2π 2 3
rz 3
= 3∫ ∫ r z+
3
dr dθ
OM
0 0
3 0
2π 2
= 3∫ ∫ ( 3r + 9r ) dr dθ
3
C
0 0
2π 2
3r 4 9r 2
S.
= 3∫ + dθ
0
4 2 0
2π
= 90 ∫ dθ
0
N
IA
2π
= 90 θ 0
= 180π
AL
Thus
→ ^
∫ F • n dS = 180π
U
S
RT
Q 2. Determine the Fourier Series of the
VI
periodic function f ( x ) shown in
the figure.
Solution:
First we will find how the periodic function in the figure be defined.
⎧π −π ≤ x ≤ 0
f ( x) = ⎨
⎩x 0≤ x ≤π
f ( x) = f ( x + 2π )
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As we know, Fourier Series is of the form
a0 ∞
f ( x) = + ∑ {an cos nx + bn sin nx}
2 n =1
where n is a positive int eger
Before doing calculation for Fourier co-efficients, its better to find whether the
given function is even or odd. As then, you know we can apply some known
results and reduce our work. If the function is odd, all the Fourier co-effiients an
for n = 0, 1, 2… are zero. If the function is even, all the Fourier co-efficients bn
OM
for n = 0, 1, 2… are zero.
C
Here, even by looking at the figure we can easily say that the given function is
S.
neither even nor odd as it is not symmetric about y-axis or origin.
Fourier co-efficients
N
IA
π
1
a0 =
π ∫π
−
f ( x) dx
AL
Put values
π π
1⎛ ⎞
0
1
a0 =
π ∫ f ( x) dx = ⎜ ∫ f ( x) dx + ∫ f ( x) dx ⎟
π ⎝ −π ⎠
U
−π 0
π
1⎛ ⎞
0
RT
= ⎜ ∫ π dx + ∫ x dx ⎟
π ⎝ −π 0 ⎠
π
1 1 x2
= π x −π +
0
VI
π π 2 0
π 3π
=π + =
2 2
π
1
an =
π ∫π
−
f ( x) cos nx dx
π π
1⎛ ⎞
0
1
an =
π ∫
−π
f ( x) cos nx dx = ⎜∫
π ⎝ −π
f ( x) cos nx dx + ∫ f ( x) cos nx dx ⎟
0 ⎠
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π
1⎛ ⎞
0
= ⎜∫ π cos nx dx + ∫ x cos nx dx ⎟
π ⎝ −π 0 ⎠
π
1 sin nx ( − cos nx )
0
sin nx
1
= π + x −
π n −π π n n2 0
1 ⎛ (−1) n 1 ⎞
= (0 − 0) + ⎜ 0 + 2 − 0 − 2 ⎟
π ⎝ n n ⎠
(−1) n − 1
=
π n2
π
1
OM
bn =
π ∫
−π
f ( x) sin nx dx
π π
1⎛ ⎞
0
1
bn = ∫ f ( x) sin nx dx = ⎜ ∫ f ( x) sin nx dx + ∫ f ( x) sin nx dx ⎟
C
π −π
π ⎝ −π 0 ⎠
S.
π
1⎛ ⎞
0
= ⎜ ∫ π sin nx dx + ∫ x sin nx dx ⎟
π ⎝ −π 0 ⎠
π
1 (− cos nx) ( − sin nx )
= π
N
(− cos nx)
1
+ x
0
−
π n π n n2
IA
−π 0
1 ⎛ −π −π (−1) n ⎞ 1 ⎛ π (−1) n +1 ⎞
= ⎜ − ⎟+ ⎜ − 0 + 0 − 0⎟
π⎝ n n ⎠ π⎝ n ⎠
AL
1 ⎛ −π + π (−1) n ⎞ 1 ⎛ π (−1) n +1 ⎞
= ⎜ ⎟+ ⎜ ⎟
π⎝ n ⎠ π⎝ n ⎠
U
n +1
−1 + (−1) (−1)
n
−1 + (−1) n + (−1) n (−1)
= + =
n n n
RT
−1 + (−1) − (−1)
n n
−1
= =
n n
VI
Hence
a0 ∞
f ( x) = + ∑ {an cos nx + bn sin nx}
2 n =1
∞ ∞
a
f ( x) = 0 + ∑ an cos nx + ∑ bn sin nx
2 n =1 n =1
3π ∞ (−1) n − 1 ∞
sin nx
f ( x) = +∑ cos nx − ∑
4 n =1 πn 2
n =1 n
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Now if n is odd
(−1) n − 1 −2
= 2
π n2 πn
and if n is even
(−1) n − 1
=0
π n2
So
3π ⎛ 2 2 2 ⎞
f ( x) = + ⎜ − cos x + (0) cos 2 x − 2 cos 3x + (0) cos 4 x − 2 cos 5 x − − − − ⎟ +
OM
4 ⎝ π 3π 5π ⎠
⎛ sin 2 x sin 3 x ⎞
⎜ − sin x − − − − − −⎟
⎝ ⎠
C
2 3
3π 2 ⎛ cos 3x cos 5 x ⎞ ⎛ sin 2 x sin 3 x ⎞
= − ⎜ cos x + + + − − − − − − − ⎟ − ⎜ sin x + + + − − − − −⎟
S.
4 π⎝ 32
5 2
⎠ ⎝ 2 3 ⎠
or
N
IA
3π 2 ∞ cos(2n − 1) x ∞ sin nx
f ( x) = − ∑ −∑
4 π n =1 (2n − 1) 2 n =1 n
U AL
RT
VI
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