CHAPTER 5 VECTOR INTEGRATION INTEGRAL THEOREMS AND Here projection is upon the xz- plane. We could also have projected on the yz-plane xy z but not on the since the surface has no component.   16 x y 2 2  xz xy 4 4 z0 4 5 Hence xzxy S S 5 x0 z0 Z Example 5.14 S Evaluate where is the surface of the F nˆ dS  S nˆ xy- sphere above the plane and    x y z a 2 2 2 2 dS .   2zkˆ F xiˆ yˆj S Y x 2  y 2  a 2 , z  0 X S xy- R Projection of on the plane is the region bounded by the circle .   ,  0 x y a z 2 2 2  2  2zkˆ   yˆj yˆj zkˆ     nˆ a 4 dS  a  ˆj dA  a dA  dz z 2 2 2 a a dA 2z  x2 y 2 2 F nˆ dS   2z x2 y a 2 2 z a z S  S S In spherical polar coordinates sin  cos, sin  sin , cos sin  x  a y  a z  a and dA  a d d 2
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Here projection is upon the xz- plane. We could also have projected on the yz-plane but not on the xy since the surface x 2  y 2  16 has no z component. 4 xz  xy 4 Hence S S xzxy 4 5 x0 z0 z0 5 Example 5.14 Z Evaluate F  nˆ dS where S is the surface of the S sphere x 2  y 2  z 2  a 2 above the xy- plane and F  xiˆ  yˆj 2zkˆ . X nˆ dS S Y x 2  y 2  a 2 , z  0 Projection of S on the xy- plane is the region R bounded by the circle x 2  y 2  a 2 , z  0 . nˆ    2 yˆj  2zkˆ 4    yˆj  zkˆ a dS   ˆj dA  a dA  a z dz a a 2 2 2 F  nˆ dS  S S x2  y 2 2z 2 a a z S x2  y 2 2z 2 z dA In spherical polar coordinates x  a sin  cos, y  a sin  sin , z  a cos and dA  a 2 sin  d d