Calc III Vector Fields: Finding Conservativity and Potential Function

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Homework Statement



Show that the vector field given is conservative and find its potential function.

F(x, y, z)= (6x^2+4z^3)i +(4x^3y+4e^3z)j + (12xz^2+12ye^3z)k.



Homework Equations





The Attempt at a Solution


When I take partial derivative with respect of y for (6x^2+4z^3) and partial derivative with respect of x (4x^3y+4e^3z) they are not equal, then is not conservative. I don't need to check the rest of them , b/c if one fail then it's not concervative.

What about the potential function? I don't think i have to do anything any more in this problem.

I will appreciate any help. Thank you.
 
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Yes, that's true. This vector field is NOT conservative and so does not have a potential function.

Notice what would happen if you tried to find a potential function:
You are looking for a function, F(x,y,z) such that
\nabla F= \frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}+ \frac{\partial F}{\partial x}\vec{k}= (6x^2+ 4z^3)\vec{i}+ (4x^3y+4e^{3z})\vec{j}+ (12xz^2+12ye^{3z})\vec{k}

So we must have
\frac{\partial F}{\partial x}= 6x^2+ 4z^3
\frac{\partial F}{\partial y}= 4x^3y+ 4e^{3z}
\frac{\partial F}{\partial z}= 12xz^2+ 12ye^{3z}

From the first of those, F(x,y,z)= 3x^3+ 4xz^3+ g(y,z) since the "constant of integration" may be a function of y and z. Differentiating that with respect to y,
\frac{\partial F}{\partial y}= \frac{\partial g}{\partial y}= 4x^3y+ 4e^{3z}
but that's impossible since the left side is a function of y and z only while the right side depends on x.

I have to ask: are you sure you have copied the problem correctly? The problem asks "Show that the vector field given is conservative", not "determine whether or not it is conservative". If the coefficient of \vec{i} were 6x^2y^2+ 4z^3, then it would be conservative.
 
Thank you so much. I found a mistake, it is conservative and I found the potential function.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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