<s>[INST] Independence of path in a closed curve line integral 
Let $f(t)$ be a continuous function. Let $C$ be a smooth closed curve. Show that $$\oint\limits_C xf(x^2 + y^2)\,dx + y f(x^2 + y^2)\,dy = 0$$ Hint: Remember that $f(t)$ has a primitive function $F(t)$. Use this fact to construct a potential function for the vector field.

If we prove that the vector field $F(x,y) = (x \cdot f(x^2 + y^2),y \cdot f(x^2 + y^2))$ is conservative in its domain, then we can use the fact $$\int\limits_C {F \cdot dr} = \varphi (r(b)) - \varphi (r(a)) = 0$$ to express the initial line integral as a potential function difference between two points on a closed smooth curve, which would imply that any path taken between these two points would yield the same integral value.
Now in order to show that the given vector field is conservative we need to find its potential, which means we have to calculate $$\frac{d\varphi}{dx} = x \cdot f(x^2 + y^2)\text{ and }\frac{d\varphi}{dy} = y \cdot f(x^2 + y^2)$$ How can we find a potential function $\varphi (x,y)$ if we do not know the form of the function $f(x^2 + y^2)$? If we replace the argument with, say, $t = {x^2} + {y^2}$ we would need to calculate the integrals $$\frac{d\varphi}{dx} = x \cdot f(t)\text{ and }\frac{d\varphi}{dy} = y \cdot f(t)$$ but we cannot treat $f(t)$ as constant since its dependent on both variables $x$ and $y$.
What i had in mind is to show that due to symmetry $F(x,y) = F(y,x)$, which implies that a closed loop is cut evenly by two curves ${C_1}$ and ${C_2}$, their sum defines the closed loop region. $F$ is conservative vector of the field if $$\frac{d}{dy} (x \cdot f(x^2 + y^2)) = 2xyf' = \frac{d}{dx} ( y \cdot f(x^2 + y^2))$$
Now we know that the field is conservative, which implies $F = \nabla \varphi $ for some scalar potential function $\varphi $ defined over the closed loop. Therefore, $$F \cdot dr = \left( \left( \frac{d\varphi}{dx} \right)i + \left( \frac{d\varphi}{dy} \right)j \right) \cdot \left( {dxi + dyj} \right) = \frac{{d\varphi }}{{dx}}dx + \frac{{d\varphi }}{{dy}}dy = d\varphi $$ Since $C$ is a continuous smooth, closed curve, parametrized, say, by $r = r(t),\,\,\,a \le t \le b$ then $r(a) = r(b)$ and $$\int\limits_C F \cdot dr = \int\limits_a^b \frac{d\varphi (r(t))}{dt} \, dt = \varphi (r(b)) - \varphi (r(a)) = 0$$ 
However, i don't think that it is sufficient to claim that the given vector field is conservative based on the equality of partial derivative of its components. We still need to find a potential function i think. For example, if instead of $F(x,y) = (x \cdot f(x^2 + y^2),y \cdot f(x^2 + y^2))$ we were given a vector field, say, $F(x,y) = (x \cdot ({x^2} + {y^2}),y \cdot (x^2 + y^2))$ we could easily find its potential to be $\varphi (x,y) = \frac{x^4}{4} + \frac{x^2 y^2}{2} + \frac{y^4}{4} = \frac{1}{4} (x^2 + y^2 )^2$. Now we can take any smooth closed curve, parametrize it and show that the potential at any two points on the curve is the same, hence its difference is zero. For example, in our case we can take a unit circle and parametrize it as $x = \cos t \text{ and }y = \sin t$. Now we just chose two random points on the curve, say, $P_1(1,0) \text{ and } P_2 (0,1)$ and show that $$\oint\limits_C x(x^2 + y^2)\,dx + y(x^2 + y^2)\,dy = \left[ \frac{1}{4} ( x^2 + y^2)^2 \right]_{(1,0)}^{(0,1)} = \frac{1}{4} - \frac{1}{4} = 0$$
How to deal when we are given $f(x^2 + y^2)$ instead of a function itself?
 [/INST] Let $g(u) := \frac{1}{2}\int_0^u f(t)\, dt$. Then $$\frac{\partial}{\partial x} g(x^2 + y^2) = \frac{1}{2}f(x^2 + y^2) \frac{\partial}{\partial x}(x^2 + y^2) = xf(x^2 + y^2)$$
and similarly $$\frac{\partial}{\partial y}g(x^2 + y^2) = yf(x^2 + y^2).$$ Therefore, the vector field $xf(x^2 + y^2)\vec{i} + yf(x^2 + y^2)\vec{j}$ is the gradient of $g(x^2 + y^2)$. Hence, $$\oint_C xf(x^2 + y^2)\, dx + yf(x^2 + y^2)\, dy = 0.$$
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