Spring 2008 Mth301 7 Solution

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Solution Assignment # 7 Let V be the region bounded by the surface x 2 + y 2 = 4 and the planes

Q 1.

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

^

where n denote the unit outer normal to S.

ts

Solution:

ud

en

Divergence Theorem or Gauss’s Theorem states that

ht Be tp: st //vu W ja eb nn si at te .n To ing H .co el m p VU St

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 ^

whose surface is S). Also suppose that n denote the unit outer normal to S at any →

point (x, y, z). If F is a vector function that has continuous partial derivatives on →

V then the volume integral of the divergence of F

i-e

→ ⎛ ⎞ ⎜ ∇ • F ⎟ over V and the ⎝ ⎠

→

surface integral of F over the surface S of V are related by →

→

^

∫ F • n dS = ∫ ∇ • F dV S

V

Or →

→

^

∫ 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 ^⎞ j+ k ⎟ • ⎜ x i+ y j+ z k ⎟ ∇ • F = ⎜ i+ ∂z ⎠ ⎝ ⎠ ⎝ ∂x ∂y 3 3 3 ∂x ∂y ∂z = + + ∂x ∂y ∂z

= 3x 2 + 3 y 2 + 3z 2 = 3( x 2 + y 2 + z 2 ) So by Divergence Theorem →

→

^

∫ F • n dS = ∫ ∇ • F dV = 3∫ ( x 2 + y 2 + z 2 ) dV V

ts

V

en

S

ud

Use cylindrical co-ordinates to evaluate the volume integral at right hand side.


x = r cos θ y = r sin θ z=z

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

Limits for

z is 0 to 3

r is 0 to 2 θ is 0 to 2π

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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→

^

2 2 2 ∫ F • n dS = ∫ ∇ • F dV = 3∫ ( x + y + z ) dV V

V

3

0

0

2

3

∫ ∫

0

2π

(r 2 cos 2 θ + r 2 sin 2 θ + z 2 ) r dz dr dθ

∫ ∫ (r

= 3∫ 0

2π

= 3∫

0

0

2

3

0

0

∫ ∫

0

2π

2

= 3∫

2π

(r 3 + rz 2 ) dz dr dθ

2

∫ ( 3r

0

3

dr dθ

3

0

= 3∫

(cos 2 θ + sin 2 θ ) + z 2 ) r dz dr dθ

rz 3 r z+ 3

∫

0

2

3

0

+ 9r ) dr dθ

0

2π

3r 4 9r 2 + 4 2

= 3∫

dθ


0

2

ts

= 3∫

2

en

2π

ud

S

2π

= 90 ∫

0

dθ

0

= 90 θ

2π

0

= 180π

Thus →

^

∫ F • n dS = 180π S

Q 2.

Determine the Fourier Series of the periodic function f ( x ) shown in the figure.

Solution:

First we will find how the periodic function in the figure be defined. ⎧π f ( x) = ⎨ ⎩x f ( x) = f ( x + 2π )

−π ≤ x ≤ 0 0≤ x ≤π


http://vujannat.ning.com Best Website To Help VUStudents As we know, Fourier Series is of the form a0 ∞ + ∑ {an cos nx + bn sin nx} 2 n =1 where n is a positive int eger f ( x) =

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

en

ts

for n = 0, 1, 2… are zero.

ud

Here, even by looking at the figure we can easily say that the given function is


neither even nor odd as it is not symmetric about y-axis or origin.

Fourier co-efficients a0 =

1

π

π

∫π

f ( x) dx

−

Put values a0 =

1

π

π

∫

−π

0 π ⎞ 1⎛ f ( x) dx = ⎜ ∫ f ( x) dx + ∫ f ( x) dx ⎟ π ⎝ −π 0 ⎠ 0 π ⎞ 1⎛ = ⎜ ∫ π dx + ∫ x dx ⎟ π ⎝ −π 0 ⎠

=

1

π

π x −π

=π + an = an =

1

π 1

π

π 2

=

0

1 x2 + π 2

π

0

3π 2

π

∫π

f ( x) cos nx dx

−

π

∫

−π

f ( x) cos nx dx =

π 0 ⎞ 1⎛ + f x nx dx f ( x) cos nx dx ⎟ ( ) cos ⎜∫ ∫ π ⎝ −π 0 ⎠



=

0 π ⎞ 1⎛ + cos π nx dx x cos nx dx ⎟ ⎜∫ ∫ π ⎝ −π 0 ⎠

sin nx 1 sin nx ( − cos nx ) = π + x − π n −π π n n2 0 1

0

π

⎛ (−1) n 1 ⎞ (0 − 0) + ⎜ 0 + 2 − 0 − 2 ⎟ π n n ⎠ ⎝ (−1) n − 1 = π n2 =

π 1

π

π

∫

f ( x) sin nx dx

−π

π

∫

−π

0 π ⎞ 1⎛ f ( x) sin nx dx = ⎜ ∫ f ( x) sin nx dx + ∫ f ( x) sin nx dx ⎟ π ⎝ −π 0 ⎠

ts

bn =

1

en

bn =

1


ud

0 π ⎞ 1⎛ = ⎜ ∫ π sin nx dx + ∫ x sin nx dx ⎟ π ⎝ −π 0 ⎠

(− cos nx) 1 (− cos nx) ( − sin nx ) = π + x − π π n2 n n −π 0 1

=

0

⎞ 1 ⎛ −π −π (−1) n ⎞ 1 ⎛ π (−1) n +1 − − 0 + 0 − 0⎟ ⎜ ⎟+ ⎜ π⎝ n n n ⎠ π⎝ ⎠

1 ⎛ −π + π (−1) n ⎞ 1 ⎛ π (−1) n +1 ⎞ = ⎜ ⎟+ ⎜ ⎟ π⎝ n n ⎠ π⎝ ⎠ n n +1 −1 + (−1) (−1) −1 + (−1) n + (−1) n (−1) = + = n n n n n −1 + (−1) − (−1) −1 = = n n

Hence a0 ∞ + ∑ {an cos nx + bn sin nx} 2 n =1 ∞ ∞ a f ( x) = 0 + ∑ an cos nx + ∑ bn sin nx 2 n =1 n =1 f ( x) =

f ( x) =

∞ 3π ∞ (−1) n − 1 sin nx +∑ cos − nx ∑ 2 πn 4 n =1 n n =1


π

http://vujannat.ning.com Best Website To Help VUStudents 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 ⎞ + ⎜ − cos x + (0) cos 2 x − 2 cos 3x + (0) cos 4 x − 2 cos 5 x − − − − ⎟ + 4 ⎝ π 3π 5π ⎠ sin 2 x sin 3 x ⎛ ⎞ − − − − −⎟ ⎜ − sin x − 2 3 ⎝ ⎠

or

ud

3π 2 ⎛ cos 3 x cos 5 x sin 2 x sin 3 x ⎞ ⎞ ⎛ − ⎜ cos x + + + − − − − − − − ⎟ − ⎜ sin x + + + − − − − −⎟ 2 2 3 4 π⎝ 3 5 2 ⎠ ⎝ ⎠


=

en

ts

f ( x) =

f ( x) =

3π 2 ∞ cos(2n − 1) x ∞ sin nx − ∑ −∑ 4 π n =1 (2n − 1) 2 n n =1


Spring 2008 Mth301 7 Solution

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