保守场与势函数

The class of fields where paths stop mattering

A worked potential computation

Recovering the potential

Why gradients deserve the name

Using $V = x^2 + y^2$, what is the work of $\nabla V$ from $(1, 0)$ to $(0, 1)$ along any path?

For $\mathbf{F} = (2x y,\ x^2)$, integrating the first component in $x$ gives $V = x^2 y + g(y)$. What comes next?

Rebuilding a potential for $(-y, x)$ leaves $\partial_y(-xy + g(y)) = -x + g'(y)$ against the needed $x$. What does this prove?

For $\mathbf{F} = (2x,\ 2y)$, give the potential $V$ with zero constant, in the form a^2+b^2.

A field has potential $V = \sin(xy)$, and work is computed from $(1, 0)$ to $(1, \pi/2)$. Give the work, as a plain number.