Verified answer

Q1

The divergence theorem states that

∫∫ F n dS

= ∫∫∫ divF dV

Where the first integral is the surface integral over the surface which bounds the volume in the second integral

For the region of this problem

= ∫0->2 ∫0->1 ∫0->3 divF dxdydz

= ∫0->2 ∫0->1 ∫0->3 ∂(x²+y)/∂x + ∂(z²)/∂y + ∂(e^y+ z)/∂z dxdydz

= ∫0->2 ∫0->1 ∫0->3 2x + 1 dxdydz

= ∫0->2 ∫0->1 (3² + 3) - (0² + 0) dydz

= ∫0->2 ∫0->1 12 dydz

= ∫0->2 12·(1-0) dz

= ∫0->2 12· dz

= 12·(2-0)

= 24

Q2

Your function is a square wave.

with amplitude 3 and period π.

The coefficients of the cosine terms are

a_k = (1/π)· ∫-π->π f(x) · cos(k·π·(x/π) dx

= (1/π) · ( ∫-π->0 (-3) · cos(k·x) dx + ∫0->π 3 · cos(k·x) dx)

= (3/π) · ( ∫-π->0 - cos(k·x) dx + ∫0->π cos(k·x) dx)

= (3/π) · (1/k) · [ (-sin(k·0) - (-sin(-k·π) + sin(k·π) - sin(k·0) ]

= 0

The coefficients of the cosine terms are

b_k = (1/π)· ∫-π->π f(x) · sin(k·π·(x/π) dx

= (1/π) · ( ∫-π->0 (-3) · sin(k·x) dx + ∫0->π 3 · sin(k·x) dx )

= (3/π) · ( ∫-π->0 -sin(k·x) dx + ∫0->π sin(k·x) dx )

= (3/π) · (1/k) · [ (cos(k·0) - cos(-k·π) - cos(k·π) +cos(k·0) ]

= (3/π) · (1/k) · [ 2 - 2·(-1)^k] for k = 1,2,3,4,5...

(For even k b_k =0)

= (12/π) · (1/k) · k = 1,3,5,7,9...

= (12/π) · (1/(2n+1)) for k = 0,1,2,3,4,5...

The Fourier series of the function is

f(x) = (4A / π) ·

∞

Σ sin((2n+1) · x) / (2n+1)

n=0