- #1
Niruli
- 2
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Hi everyone! I've been having a hard time figuring this one out for a while, so any help will be appreciated!
[itex]\textbf{F}[/itex]= <(2zx)/(x^2+e^z*y^2), (2ze^z*y)/(x^2+e^z*y^2), log(x^2+e^z*y^2) + (ze^z*y^2)/(x^2+e^z*y^2)>
(a) Where is the following vector field defined?
(b) Is this region simply connected?
(c) Find a potential function for [itex]\textbf{F}[/itex].
[itex]\textbf{F}[/itex]= ∇V, where V is the potential function.
I answered (a) and (b). For (c), I computed the x- and y-components of V by integrating the x- and y-components of [itex]\textbf{F}[/itex], but I have no idea how to integrate the z-component of [itex]\textbf{F}[/itex] to find the z-component of V.
dV/dx= ∫(2zx)/(x^2+e^z*y^2)dx= zln(x^2+e^z*y^2) + f(y,z)
dV/dy= ∫(2ze^z*y)/(x^2+e^z*y^2)dy= zln(x^2+e^z*y^2) + g(x,z)
dV/dz= ∫(log(x^2+e^z*y^2) + (ze^z*y^2)/(x^2+e^z*y^2))dz
Homework Statement
[itex]\textbf{F}[/itex]= <(2zx)/(x^2+e^z*y^2), (2ze^z*y)/(x^2+e^z*y^2), log(x^2+e^z*y^2) + (ze^z*y^2)/(x^2+e^z*y^2)>
(a) Where is the following vector field defined?
(b) Is this region simply connected?
(c) Find a potential function for [itex]\textbf{F}[/itex].
Homework Equations
[itex]\textbf{F}[/itex]= ∇V, where V is the potential function.
The Attempt at a Solution
I answered (a) and (b). For (c), I computed the x- and y-components of V by integrating the x- and y-components of [itex]\textbf{F}[/itex], but I have no idea how to integrate the z-component of [itex]\textbf{F}[/itex] to find the z-component of V.
dV/dx= ∫(2zx)/(x^2+e^z*y^2)dx= zln(x^2+e^z*y^2) + f(y,z)
dV/dy= ∫(2ze^z*y)/(x^2+e^z*y^2)dy= zln(x^2+e^z*y^2) + g(x,z)
dV/dz= ∫(log(x^2+e^z*y^2) + (ze^z*y^2)/(x^2+e^z*y^2))dz