CHAPTER 4 GRADIENT, DIVERGENCE CURL AND 1 2 So that 2 4z   x  y   2 holding y and z constant. Similarly, 2x 3y  2xy    z y yz gx, z 32 2  4x  2z   4xz   y yz z hx, y 2 Comparing the three equations, we have a common value of  if we choose x2 , ,   z gx, z z hx, y 3y 2 2 y, 2 2  f z 2  2 2  2 2 2xy 4xz      z yz 2 2 2 So that 2 2 DIVERGENCE: (x,y,z) Let be defined and differentiable at every point in a    Ax, y, z A1iˆ A2 A3kˆ ˆj specified region. Then the divergence of , written or is defined by A  A div A ˆ  ˆ  i    We note that the gradient operator acts on a scalar and gives a vector while divergence operation acts on a vector and gives a scalar. Example 4.10 ˆj If , determine .  2x 2xy   3 A yiˆ y z y z kˆ  A 2 2 3 2 2  i ˆ  ˆ  4xy 4xy 6  6  0 y z y z 2 2
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL So that  x  2 y  4z   1 2 2 holding y and z constant. Similarly,  2x 3y z   2xy 32 y 2 yz  gx, z  4x y  2z   4xz yz  z 2  hx, y Comparing the three equations, we have a common value of  if we choose f y, z  3y 2 2  z 2 , gx, z  x2 2  z 2 , hx, y  2 2 2 2 So that   2 2 2 2  z 2  2xy  4xz yz DIVERGENCE: Let Ax, y, z  A1iˆ  A2 ˆj  A3kˆ be defined and differentiable at every point (x,y,z) in a specified region. Then the divergence of A , written  A or div A is defined by i  ˆ  ˆ    We note that the gradient operator acts on a scalar and gives a vector while divergence operation acts on a vector and gives a scalar. Example 4.10 If A  2x 2 yiˆ 2xy 2  y 3 z ˆj  3 y 2 z 2 kˆ , determine  A . i  ˆ  ˆ  4xy 4xy 6 y 2 z  6 y 2 z  0