Begin2.DVI

To evaluate the line integral let

x= cos θ, y = sin θwith dx =−sin θ dθ, dy = cos θ dθ

and show

I 4 =

∫

C

F
·d
r =

∫

C

(y−z)dx + (2 x−y)dy =

∫ 2 π

0

[−sin^2 θ+ 2 cos^2 θ−sin θcos θ]dθ =π

Note that z= 0 on C.

Example 7-38. Evaluate the flux integral I=

∫∫

S

F
·dS
where the vector field

is given by F
=F
(x, y, z ) = zeˆ 1 +yˆe 2 +xeˆ 3 and S is the surface of the unit sphere

x^2 +y^2 +z^2 = 1.

Solution Transform to spherical coordinates where the position vector to a point of

the unit sphere is

r =xˆe 1 +yˆe 2 +zˆe 3 = sin θcos φˆe 1 + sin θsin φˆe 2 + cos θˆe 3

for 0 ≤φ≤ 2 πand 0 ≤θ≤π. An element of surface area in spherical coordinates is

dS = sin θ dθdφ and a unit normal ˆento the surface of the unit sphere is in the same

direction as the vector
r above so that one can write

ˆen= sin θcos φeˆ 1 + sin θsin φˆe 2 + cos θeˆ 3

Substituting these values into the flux integral one obtains

I=

∫∫

S

F
·ˆendS =

∫ 2 π

φ=0

∫π

θ=0

[cos θ(sin θcos φ) + (sin θsin φ)(sin θsin φ) + (sin θcos φ) cosθ] sin θ dθdφ

Evaluating the inner integral, holding φconstant gives

I=

∫ 2 π

0

4

3 sin

(^2) φ dφ

which can be integrated to obtain the value I=^4 π

3

.