1. Aug 1, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

$\nabla{F} = <2xyze^{x^2},ze^{x^2},ye^{x^2}$
if f(0,0,0) = 5 find f(1,1,2)

2. Relevant equations

3. The attempt at a solution

my book doesn't have a good example of a problem like this, am I looking for a potential?

$<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} >= <2xyze^{x^2},ze^{x^2},ye^{x^2}>$

Last edited: Aug 1, 2014
2. Aug 1, 2014

### CAF123

Yes, solve the three differential equations to find F(x,y,z) up to a constant. The given condition sets this constant, from which you can find F(x,y,z) for all x,y,z.

3. Aug 1, 2014

### jonroberts74

$\frac{\partial}{\partial x} = 2xyze^{x^2} \Rightarrow \int 2xyze^{x^2}dx = yze^{x^2} + h(y,z)$

$\frac{\partial}{\partial y} = ze^{x^2} = ze^{x^2} + h(y,z) \Rightarrow h_{y}(y,z) = 0$

$\frac{\partial}{\partial z} = ye^{x^2} = ye^{x^2}+h(z) \Rightarrow h_{z}(z) = h'(z)$

$h'(z) = 0 \Rightarrow h(z) = k$ for a constant k, and because f(0,0,0) = 5 then

$f= yze^{x^2} + 5$

so $f(1,1,2) = 1(2)e^1+5 = 2e+5$

Last edited: Aug 1, 2014
4. Aug 1, 2014

Correct!

ehild