Mathematical Physics Vol 1

Chapter 4. Field theory

180

Exercise 122

Calculate

y V

∇ · A d V , where A = 4 x i − 2 y 2 j + z 2 k is a vector field observed over the region V , bounded by surfaces x 2 + y 2 = 4, z = 0 and z = 3: a) directly, b) by using the divergence theorem (Gauss’s theorem). Solution a) The volume V is the volume of a cylinder obtained when the unbounded cylinder x 2 + y 2 = 4 is intersected by planes z = 0 and z = 3, and hence the volume integral is I = y V ∇ · A d V = y V ∂ ∂ x ( 4 x )+ ∂ ∂ y ( − 2 y 2 )+ ∂ ∂ z ( z 2 ) d V = = y V ( 4 − 4 y + 2 z ) d V = 3 Z 0 ( 4 − 4 y + 2 z ) d z = 84 π . b) The volume V is bounded by the surface S , composed of the lower base S 1 ( z = 0), upper base S 2 ( z = 3) and side S 3 ( x 2 + y 2 = 4). Applying the divergence theorem, we obtain the surface integral I = I 1 + I 2 + I 3 = x S A · n d S = = x S 1 A · n d S 1 + x S 2 A · n d S 2 + x S 3 A · n d S 3 . = 2 Z − 2 d x √ 4 − x 2 Z − x 2 − √ 4 d y

Let us now calculate the three integrals. For S 1 ( z = 0), n = − k , A = 4 x i − 2 y

2 j and A · n = 0, and the integral is

I 1 = x S 1

A · n d S 1 = 0 .

2 j + 9 k and A · n = 0, and the integral is

For S 2 ( z = 3), n = k , A = 4 x i − 2 y

I 2 = x S 2

A · n d S 2 = 9 x S 2

d S 2 = 36 π ,

because the area of S 2 is equal to 4 π . For S 3 ( x

2 + y 2 = 4) the unit vector of the normal is

∇ ( x 2 + y 2 − 4 ) | ∇ ( x 2 + y 2 − 4 ) |

2 x i + 2 y j p 4 x 2 + 4 y 2 x i + y j 2

x i + y j 2

n =

=

=

A · n =( 4 x i − 2 y 2 j + z 2 k ) ·

= 2 x 2 − y 3 .