) If F(x, y, z) = e xˆi + yzˆj − yz2ˆk, find the divergence of F at (0, 2, −1)
If "F(x, y, z) = e^x\\cdot i + yz\\cdot j \u2212 yz^2\\cdot k", let us find the divergence of "F" at "(0, 2, \u22121)."
It follows that
"Div\\ [F(x,y,z)]=\\frac{\\partial( e^x)}{\\partial x}\n+\\frac{\\partial(yz)}{\\partial y}\n+\\frac{\\partial(-yz^2)}{\\partial z}\n=e^x+z-2yz,"
and hence
"Div\\ [F(0,2,-1)]=e^0-1-2\\cdot 2(-1)=4."
Answer: the divergence of "F" at "(0, 2, \u22121)" is equal to 4.
Comments
Leave a comment