How to calculate the line integral when the vector field is conservative?

Question

Calculate $$\int_{\gamma} x^3 dx + (x^3 + y^3)dy$$ where $\gamma$ is the line $y=2-x$ from $(0, 2)$ to $(2, 0)$.

It was my understanding that if the vector field was conservative then the path doesn't matter so what I did was check if the vector field was conservative and yes it was conservative since I found a potential function to it which was: $$f = \frac{x^4}4 + \frac{y^4}4$$

then $$\int_{\gamma} x^3 dx + (x^3 + y^3)dy = f(2,0) - f(0, 2) = 0$$

This is wrong and the correct answer is -4.

Answer 1

Parametrizing the curve you get:

$$ \gamma(t)=(t,2-t) \quad \text{with} \quad t\in[0,2] \quad\text{and} \quad\,\gamma'(t)=(1,-1)$$

$$\int_{\gamma} x^3dx+(x^3+y^3)dy=\int_0^2(t^3-t^3-(2-t)^3)dt=-\int_0^2 u^3 du=-4 $$ In this case, since $ \dfrac{\partial }{\partial y}x^3\neq \dfrac{\partial }{\partial x}(x^3+y^3) $ the vector field is not conservative.

Answer 2

The potential $f = \frac{x^4}4 + \frac{y^4}4$ gives just a part of the given field $\mathbf{F}=(x^3,x^3+y^3)$. Indeed $\nabla f (x,y)=(x^3,y^3)$ and the remaining part $\mathbf{F}-\nabla f (x,y)=(0,x^3)$ is not conservative because it is not irrotational: $ \frac{\partial }{\partial x}(x^3)-\frac{\partial }{\partial y}(0)\not=0$. Hence, we may evaluate the line integral by splitting $\mathbf{F}$ in two, $$\begin{align}\int_{\gamma} (x^3,x^3+y^3)\cdot d\mathbf{s}&=\int_{\gamma} (x^3,y^3)\cdot d\mathbf{s}+\int_{\gamma} (0,x^3)\cdot d\mathbf{s}\\ &=f(2,0) - f(0, 2) +\int_{\gamma} (0,x^3)\cdot d\mathbf{s}\\ &=4-4+\int_{2}^0 (0,t^3)\cdot (-1,1)\,dt\\ &=0+\left[\frac{t^4}{4}\right]_2^0=-4 \end{align}$$ where $\gamma(t)=(2-t,t)$ with $t\in[0,2]$.