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Consider the following vector field:

$$\vec A(x,y,z)=(yz)\hat i+(xz)\hat j+(xy)\hat k$$

Compute the line integral of $A$ along a path of your choice connecting $(0,0,0)$ to $(1,1,1).$

I recognise that $\nabla\times A=0$ and so I can simply evaluate $V(1,1,1)-V(0,0,0)$ where $A=\nabla V$.

I'm not sure how to find the function $V$ though.

The solution says $V=xyz.$

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  • $\begingroup$ Added finding potential into your title. $\endgroup$ – MrYouMath May 16 '16 at 21:06
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Assume V as a function of x,y and z.

Then $\nabla V=(\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z})$

Compare components:

$$\dfrac{\partial V}{\partial x}=yz$$ $$\dfrac{\partial V}{\partial y}=xz$$ $$\dfrac{\partial V}{\partial z}=xy$$

Integrate: $$V=xyz+f_1(y,z)$$ $$V=xyz+f_2(x,z)$$ $$V=xyz+f_3(x,y)$$

From here you can see that $V=xyz+const.$ by inspection. If you want to do it mathematically you have to subtract two equations from each other.

1-2: $f_1(y,z)=f_2(x,z)$ from that you can conclude $f_1(y,z)=f_1(z)$ and $f_2(x,z)=f_2(z)$, or you would have $y$ on the left but no $y$ on the right, and so forth.

1-3: $f_1(z)=f_3(x,y)$.As one can see we have z on the left sid but no z on the right. Hence, $f_1$ is a constant function as it is not depending on $z$. if $f_1=const.$ $f_1=f_3(x,y)$ will imply that $f_3=f_1=const.$. Same argument as before and we can conclude $f_1=f_3=const$. But if $f_1=const.$ than from 1-2 $f_2=const$.

With that we obtain $f_i=const$ for $i=1,2,3$. At the end you will see, that $f_i=const$ for $i=1,2,3$.

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  • $\begingroup$ I don't understand how you've gone from $f_1(y,z)=f_2(x,z)$ to $f_1(y,z)=f_1(z).$ $\endgroup$ – Si.0788 May 16 '16 at 21:10
  • $\begingroup$ The equation tells you, that we have something with y and z on the left hand side and something with x and z on the right side of the equation. But it is not possible to have y only on one side of the equation, hence it must vanish and $f_1$ can only be a function of z. The same argument works for $f_2$. $\endgroup$ – MrYouMath May 16 '16 at 21:13
  • $\begingroup$ Ok, that makes sense. Now how do you conclude that $f_i=c$? $\endgroup$ – Si.0788 May 16 '16 at 21:15
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$\nabla G = A$

what is G?

$G = \int yz \,dx + f(y,z) = \int xz \,dy + g(x,z) = \int xz \,dz + h(x,y)\\ G = xyz + f(y,z) = xyz + g(x,z) = xyz + h(x,y)$

$f(y,z), g(x,z), h(x,y)$ must all be constant functions, and they can be any constant function, so might as well make them $0.$

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