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Your statement about whether $\vec{F}$ is a conservative field is wrong. $\vec{F}$ is conservative iff $\nabla \times \vec{F} = 0$. In terms of the Cartesian components, the curl is $$\nabla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial P}{\partial z} \right) \hat{x} + \left( \frac{\partial Q}{\partial z} - \frac{\partial ... 4 Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function f(x) to a field, which is a function of many variables f(x, y, z), we suddenly have ... 2 The work-energy theorem leads us to the following result; $$\oint \vec F\cdot d\vec s=0$$ $$\oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n$$ Using the rules of vector calculus there must exist some scalar function such that; \vec ... 1 Let F be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero$$\oint_{O} F \cdot dr = 0$$The del operator \nabla is defined in 3 dimensions as$$\nabla =\left\langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, ...
On spherical coordinates, the gradient of a general function $V$ is: $$\nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi$$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...