Hello I do not have much experience in these problems so I am looking to see if anyone can help me to look over the following;
For the vector function $F(2xy,x^2+2yz,y^2+1)$ determine if $\nabla \psi=F$ has a solution and determine $\phi$ if possible and then calculate the line integral from the origin to $(1,1,1)$ along $x=y=z$ and then along the curve $y=x^2,z=x^3$. I included my answer and I also would appreciate if someone could check if my final numerical answers are correct or if I made mistakes.
My thoughts:
Through the process of taking partial derivatives and doing reverse integration, I found that $$\psi(x,y,z)=x^2y+y^2z+z$$
I believe that this tells me that the vector field is conservative, all though I do not really have a deep understanding of what that can tell us.
For the next part,
I know that the line integral for a vector function is given as $$\int_{C} F(r(t)) \bullet r'(t) dt$$
We have $$F(x,y,z)=(2xy)i+(x^2+2yz)j+(y^2+1)k$$
I am not sure what is meant by along $x=y=z$,
does this just mean by the line segment given by the line segment
$r(t)=(t,t,t)$ , $0 \le t \le 1$
So $$F(r(t))=(2t^2)i+(3t^2)j+(t^2+1)k$$
and dotted with (1,1,1)
So we would just need to evaluate $\int_{0}^{1} 6t^2+1=3$
For $(0,0,0)$ to $(1,1,1)$ via $y=x^2$ and $z=x^3$ I am less sure.
It seems that $x$ is the parameter. But I dont know how I can find the $r(t)$. That is where I am at so far. Would it just be that $r(t)=(t,t^2,t^3)$ and so $r'(t)=(1,2t,3t^2)$ and thus would come down to evaluating the $\int_{0}^{1} 2t^3+2t^3+4t^6+3t^6+3t^2 dt=\int_{0}^{1} 3t^2+4t^3+7t^6 dt=3$
To make sure I understand the method, I also want to try evaluating over the same paths but with $$F(x,y,z)=(2xyz^3)i+(-x^2z^3-2y)j+(3x^2+yz^2)k$$
I used the same method but it seems like my answer is not the same as other people who have checked it so I must be making a mistake. I wont include all the work but here is what I tried.
Over $x=y=z$ $F(r(t))=(2t^5)i+(-t^2-2t)j+(3t^5)k$
$F(r(t)) \bullet r'(t)= 4t^5-2t$. evaluating that integral from $0$ to $1$ comes as $\frac{-1}{3}$
over the other path
I calculated $F(r(t))=(2t^{12})i+(-t^{11}-2t^2)j+(3t^{10})k$
and $F(r(t)) \bullet r'(t)= 11t^{12}-2t^{11}-4t^{3}$ which I evaluated to be $\frac{-25}{78}$
Does anyone see where I could have gone wrong? Thanks in advance
You've found a potential function $\psi$ for your vector field $F$. So you know it's a conservative field. Now you can apply the fundamental theorem of calculus for line integrals to calculate both of those line integrals without doing any actual integration.
$$ \int_C \mathbf{F \cdot dr} = \psi(b) - \psi(a) $$
where $a$ is the starting point of the path $C$ and $b$ is the endpoint of the path $C$.
Since the field is conservative, the actual path taken from the origin to $(1,1,1)$ doesn't matter - both line integrals have the same value.