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·dr =
∫
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