CHAPTER 5 VECTOR INTEGRATION INTEGRAL THEOREMS AND Taking ; nˆ  kˆ   2z n  k   Then 2 S  dA   2 2 dA 2z nˆ z  R R R xy- Now since we are in the plane, we have from that    2  2 x y z z x y 2 2 2 2 2 Hence dxdy  2 S x2 2 x y 2 2 We let , then, , where J is the    cos ,  sin  x r y r dxdy Jdrd r, , (x,y) Jacobian of with respect to i.e.  r  sin sin  r cos r Hence   dxdy rdrd  rdrd 0     Then  2 S 2 r 2 2 1 2 Example 5.10 Find the area of the surface cut from the bottom of the paraboloid by the   z x y 2 2 plane  1. z The surface projects onto the disk in the xy- plane. We apply : 1 R x  y  2 2 x, to get z  f y  x  y 2 2  Area Z   1dxdy 4r fx2 fy 2 2 4x2  1 4 y x2  2   z  x  y 2 2 1  S R0 Rxy  1 z 6 2 1
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Taking nˆ  kˆ ;  n    k  2z Then S  R  nˆ dA  R 2 2 2z  2 R dA z Now since we are in the xy- plane, we have from x 2  y 2  z 2  2 that z  2 x 2 y 2 Hence S  2 x2 dxdy 2 x 2 y 2 We let x  r cos , y  r sin  , then, dxdy  Jdrd  , where J is the Jacobian of (x,y) with respect to r, , i.e. r sin  sin  r cos  r Hence dxdy  rdrd  Then S  2 1 2 rdrd  2 r 2 2 0     Example 5.10 Find the area of the surface cut from the bottom of the paraboloid z  x 2  y 2 by the plane z  1. The surface projects onto the disk R : x 2  y 2  1 in the xy- plane. We apply z  f x, y  x 2  y 2 to get Area  fx2  fy 2 1dxdy  Rxy 2 1  4r 2 1   R0 x2  6 4x2  4 y 2 1 S Z z  1 z  x 2  y 2