how to show $\frac{\partial f(x,y,z)}{\partial x}=F_1(x,y,z)$ using the fundamental theorem of calculus

Question

If $\displaystyle f(x,y,z)= \int_c (F\,dl)=\cdots=\int_0^x (F_1(t,0,0)\,dt)+\int_0^y (F_2(x,t,0)\,dt)+\int_0^z (F_3(x,y,t)\,dt)$

Why $\dfrac{\partial f}{\partial x}=F_1(x,y,z)$ ?

My attempt: differentiating in respect to x the last terms from the above equation are zero since neither varies with respect to x,to the first term we can make $F_1(x,0,0)=G(x)$ and $$\frac{d}{dx} \int_0^x G(t)=G(x)=F_1(x,0,0)=\frac{\partial f(x,0,0)}{\partial x}$$

I just prove for f(x,0,0) how I change something so that I prove to $f(x,y,z)$ or $\partial f(x,y,z)/\partial x=F_1(x,y,z)$ ?

Thank you

Answer 1

The point is that if the field is conservative, you can instead of choosing a path $(0,0,0)\to (x,0,0)\to (x,y,0)\to (x,y,z)$ like in your post, choose a path $(0,0,0)\to (0,y,0)\to (0,y,z)\to (x,y,z)$. In this case you'll get for some function $G(y,z)$, that $$f=G(y,z)+\int_0^x F_1(t,y,z)dt$$ much like in your equation in the original post. so that $\frac{\partial f}{\partial x}=F_1$ follows immediately.

Answer 2

The reasoning is false. The second two integrals do depend on $x$, so the correct partial derivative gives:

$$\frac{\partial f(x,y,z)}{\partial x}=F_1(x,0,0)+\int_0^y\frac{\partial F_2(x,t,0)}{\partial x}dt+\int_0^y\frac{\partial F_3(x,t,0)}{\partial x}dt$$

Do you have other conditions for this problem? It looks like you're evaluating a line integral first along the $x$ axis, then along the $y$ axis, then along the $z$ axis.

I don't see how your question is related to what you're trying to prove: that two line integrals with the same endpoints have the same value if the vector field is conservative. For two line integrals between two endpoints, write them as $\int_{c_1}F(x(t),y(t),z(t))dt$, $\int_{c_2}F(x(t),y(t),z(t))dt$, where $c_1,c_2$ have the same endpoints. Notice that if you reverse $c_2$ (the integration direction along $c_2$, then

$$\int_{c_1}F(x(t),y(t),z(t))dt+\int_{-c_2}F(x(t),y(t),z(t))dt=0,$$

since you're now integrating along a loop. But reversing $c_2$ is equivalent to changing $t\rightarrow -t$, which means that $\int_{-c_2}F(x(t),y(t),z(t))dt=-\int_{c_2}F(x(t),y(t),z(t))dt$. So you get the desired conclusion.