I need help with this problem. Problem is attached. Thanks
M meks0899 New member Joined Aug 27, 2009 Messages 17 Dec 13, 2009 #1 I need help with this problem. Problem is attached. Thanks Attachments Problem 3.jpg 9.9 KB · Views: 113
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Dec 13, 2009 #2 fxx+fyy=0 ⟹ fxx=−fyy ⟹ ∂∂yfy=−∂∂xfx\displaystyle f_{xx}+f_{yy}=0 \implies f_{xx}=-f_{yy} \implies \frac{\partial}{\partial y}f_y = -\frac{\partial}{\partial x}f_xfxx+fyy=0⟹fxx=−fyy⟹∂y∂fy=−∂x∂fx By Green's theorem, ∮Cfydx+(−fx)dy=∫∫D[∂∂x(−fx)−∂∂yfy]dA\displaystyle \oint_C f_y dx + (-f_x) dy = \int \int _D \left [ \frac{\partial}{\partial x} (-f_x) - \frac{\partial}{\partial y} f_y \right] dA∮Cfydx+(−fx)dy=∫∫D[∂x∂(−fx)−∂y∂fy]dA Then you're "almost there."
fxx+fyy=0 ⟹ fxx=−fyy ⟹ ∂∂yfy=−∂∂xfx\displaystyle f_{xx}+f_{yy}=0 \implies f_{xx}=-f_{yy} \implies \frac{\partial}{\partial y}f_y = -\frac{\partial}{\partial x}f_xfxx+fyy=0⟹fxx=−fyy⟹∂y∂fy=−∂x∂fx By Green's theorem, ∮Cfydx+(−fx)dy=∫∫D[∂∂x(−fx)−∂∂yfy]dA\displaystyle \oint_C f_y dx + (-f_x) dy = \int \int _D \left [ \frac{\partial}{\partial x} (-f_x) - \frac{\partial}{\partial y} f_y \right] dA∮Cfydx+(−fx)dy=∫∫D[∂x∂(−fx)−∂y∂fy]dA Then you're "almost there."
