Question

F(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (1, 2) to (2, 8) (a) find a function f such that f = ∇f.

Answer 1

`\mathbf F(x,y)=x^2\,\mathbf i+y^2\,\mathbf j`

If a scalar function
`f(x,y)` exists for which
`\\abla f(x,y)=\mathbf F(x,y)`, then we'd have


`(\partial f)/(\partial x)=x^2\implies f(x,y)=\frac{x^3}3+g(y)`

`\implies(\partial f)/(\partial y)=y^2=g'(y)\implies g(y)=\frac{y^3}3+C`

so the scalar function would be given by


`f(x,y)=\frac{x^3+y^3}3+C`

where
`C` is an arbitrary constant.

Presumably, this question is being asked in the context of the line integral of
`\mathbf F(x,y)` along the given contour
`C`. Since
`\mathbf F(x,y)` is conservative - and consequently a scalar potential function
`f(x,y)` exists - we can simply use the gradient theorem to evaluate the line integral. We would get


`\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=f(2,8)-f(1,2)=\frac{511}3`

(and we get the same answer by parameterizing
`C` and computing the integral as we usually would)