UNIT-4
Q:1 Find the Fourier series of f(x) = x in the interval
Solution: Here 
; 
It’s Fourier series is given by
… (1)
Where
&
Hence the required Fourier series is
Q:2 Find the Fourier series for
in the interval 
Hence deduce that
Solution:
Here 
;
Hence it’s Fourier series is,
… (1)
Where
&
Hence equation (1) becomes
Put 
we get
i.e. 
Q:3 Find a Fourier series expansion in the interval 
for
;
;
Solution: Here
;
;
Hence it’s Fourier series expansion is,
… (1)
Where
And
Hence equation (1) becomes
Q:4 Find a Fourier series of
;
;
Solution:
Here
;
;
Here f(x) is odd function Hence we get half range sine series i.e. 
… (1)
Where
Hence equation (1) becomes,
Q:5 Find a Fourier series for
;
Solution:
Here
;
Since f(x) is even function hence 
It’s Fourier series is
… (1)
Where
Hence equation (1) becomes,
Q:6 Find half range cosine series of 
in the interval 
and hence deduce that
a) 
b) 
Solution:
Here
;
Hence it’s half range cosine series is,
… (1)
Where
Hence equation (1) becomes,
… (2)
Put x = 0, we get
Hence the result
Put 
we get,
i.e. 
Q:7 Using complex form, find the Fourier series of the function
f(x) = sinnx = 
Solution:
We calculate the coefficients 
= 
=
Hence the Fourier series of the function in complex form is
We can transform the series and write it in the real form by renaming as
n=2k-1,n=
= 
Q:8 Using complex form find the Fourier series of the function f(x) = x2, defined on te interval [-1,1]
Solution:
Here the half-period is L=1. Therefore, the co-efficient c0 is,
For n
Integrating by parts twice, obtain
= 
= 
= 
.
= 
.
Q:9 Find the gradient of the following:
Solution: 
y=
= 
.
= 2x+
Q:10 Find the curl of F(x,y,z) = 3
i+2zj-xk
Curl F = 
=
= 
i - 
= (0-2)i-(-1-0)j+(0-0)k
= -2i+j
Q:11 What is the curl of the vector field F= ( x +y +z ,x-y-z,
)?
Solution:
Curl F = 
= 
=
= (2y+1)i-(2x-1)j+(1-1)k
= (2y+1)i+(1-2x)j+0k
= (2y+1, 1-2x,0)
Q:12 Compute 
where F= (3x+
and s is the surface of the box such that 0
use outward normal n
Solution: Writing the given vector fields in a suitable manner for finding divergence
div F =3+2y+x
We use the divergence theorem to convert the surface integral into a triple integral
Where B is the box 0
, 0
We compute the triple integral of div F=3+2y+x over the box B
=
=
= 36+3=39
Q:13 For F= (
use divergence theorem to evaluate 
where s is the dphere of radius 3 centred at origin.
Solution: Since div F= 
, the surface integral is equal to the triple integral.
To evaluate the triple integral we can change value of variables to spherical co-ordinates,
The integral is 
=
.For spherical co-ordinates, we know that the jacobian determinant is dV = 
.therefore, the integral is
= 
= 
=
