Question

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f such that f = ∇f. f(x, y) = x3 3​+ y3 3​ correct: your answer is correct. (b) use part (a) to evaluate c ∇f · dr along the given curve
c.

Answer 1

If there is some scalar function $f(x,y)$ such that

$\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j$

then we want to find $f$ such that

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C$
$\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C$

So the vector field $\mathbf f(x,y)$ is conservative, which means the fundamental theorem applies; the line integral of $\mathbf f$ along any path $\mathcal C$ parameterized by some vector-valued function $\mathbf r(t)$ over $a\le t\le b$ is given by

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))$

In this case,

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23$