Now on the surface Sdefined by z=z(x, y )one finds F 1 =F 1 (x, y, z ) = F 1 (x, y, z (x, y ))
is a function of x and y so that a differentiation of the composite function F 1 with
respect to yproduces
∂F 1 (x, y, z (x, y ))
∂y =∂F 1
∂y +∂F 1
∂z∂z
∂y (8 .48)which is the integrand in the integral (8.47) with the sign changed. Therefore, one
can write ∫∫
S(
−∂F^1
∂z∂z
∂y−∂F^1
∂y)
dx dy =−∫∫S∂F 1 (x, y, z(x, y ))
∂ydx dy (8 .49)Now by using Greens theorem with M(x, y ) = F 1 (x, y, z (x, y )) and N(x, y ) = 0 , the
integral (8.49) can be expressed as
−∫∫S∂F 1 (x, y, z (x, y ))
∂y dx dy =∫CpF 1 (x, y, z (x, y )) dx =∫CF 1 (x, y, z)dx (8 .50)which verifies the first integral of the equations (8.44). The remaining integrals in
equations (8.44) may be verified in a similar manner.
Example 8-11. Verify Stokes theorem for the vector field
F= 3x^2 yˆe 1 +x^2 yˆe 2 +zˆe 3 ,where Sis the upper half of the sphere x^2 +y^2 +z^2 = 1.
Solution The given vector field has the curl vector
curl F =∇× F =∣∣
∣∣
∣∣ˆe 1 eˆ 2 ˆe 3
∂
∂x∂
∂y∂
∂z
3 x^2 y x^2 y z∣∣
∣∣
∣∣= (2 xy −^3 x(^2) )ˆe 3.
The unit normal to the sphere at a general point (x, y, z )on the sphere is given by
ˆen=xˆe 1 +yˆe 2 +zeˆ 3and the element of surface area dS when projected upon the xy plane is
dS = ˆedx dy
n·eˆ 3=dx dyz.