line integral of a gradient vector field

Question

I'm having trouble finding the line integral of this problem. So they give me the gradient vector field $$\nabla f(x,y,z) = 2xyze^{x^2}i + ze^{x^2}j+ye^{x^2}k$$ they say that $f(0,0,0) = 5$ and i want to know what is the value of line integral $f(1,1,2)$

i know that the line integral of a gradient vector field is given by $$\int_{a}^{b} \nabla f(x,y,z)ds = f(c(b)) -f(c(a))$$ But i'm stuck by how to find $f$, can someone explain to me

Answer 1

$$\frac{\partial f}{\partial x}=2xyze^{x^2}\implies f(x,y,z)=\int\frac{\partial f}{\partial x}\,dx= yze^{x^2}+g(y,z)$$ and $$\frac{\partial f}{\partial y}=ze^{x^2} \implies f(x,y,z)=\int\frac{\partial f}{\partial y}\,dy=yze^{x^2}+h(x,z)$$ and $$\frac{\partial f}{\partial z}=ye^{x^2} \implies f(x,y,z)=\int\frac{\partial f}{\partial z}\,dz=yze^{x^2}+m(x,y)$$ So you may take $f(x,y,z)=yze^{x^2}+5$ (choose $g,h,m=5$ to ensure $f(0,0,0)=5$) and then use the formula you know for the line integral.

Answer 2

I think that the gradient should be $$\nabla f(x,y,z)=2xyze^{x^2}i + \mathbf{z}e^{x^2}j+ye^{x^2}k$$ where the variable $z$ has been inadvertently replaced by the number $2$ (otherwise it cannot be a gradient of some function $f$). Then by integration we obtain $f(x,y,z)=e^{x^2}yz+\text{constant}$. Can you take it from here?

Answer 3

I think there's been a misunderstanding, the vector field you gave isn't conservative. I'm pretty sure the $y$ term should be $ze^{x^2}$, not $2e^{x^2}.$ Was the problem handwritten? In any case, I'm going to assume that was what you meant because otherwise the problem isn't solvable. (Use Clairaut's Theorem to figure out why)

The method I was taught to do this was to take kind of "partial integrals", like partial derivatives but in reverse. So essentially it's just like a normal integral, except you treat any other variables as constants and your constant is actually a function of the variables you hold constant.

So first, we have that $f_x = 2xyze^{x^2}$, so $f(x,y,z) = \int 2xyze^{x^2} dx = yze^{x^2} + g(y,z)$.

$g$ being a function of $y$ and $z$ is necessary because any function of $y$ and $z$ will disappear when we differentiate with respect to $x$, not just a constant.

Now that we have an expression for $f$, we can differentiate it with respect to $y$ and compare it to what we have in the gradient: $f_y = ze^{x^2} + g_y = ze^{x^2} \to g_y = 0$. This is where the problem would become impossible with the original question: we would get $g_y = (2-z)e^{x^2}$, which is impossible because $g$ is not dependent on $x$.

Now we can do our partial integration again: $g(y,z) = \int 0 dy = h(z)$. Plugging this back into our expression for $f$ yields $f(x,y,z) = yze^{x^2} + h(z)$. This step is a bit overkill here because we already see that the $y$-derivatives match, but it's included for completeness because that's not always the case.

I've excluded the last bit for brevity's sake but repeating this one more time will give you that $h$ is just a constant, so the solution is $f(x,y,z) = yze^{x^2} + C$.

Hope this helps!

Edit: I see that you've addressed the y component issue now, so sorry if it seemed like I was laboring the point. It took a bit to write this.