Question

Consider f and c below. f(x, y) = (3 + 4xy2)i + 4x2yj, c is the arc of the hyperbola y = 1/x from (1, 1) to 3, 1 3 (a) find a function f such that f = ∇f. f(x, y) = (b) use part (a) to evaluate c f · dr along the given curve
c.

Answer 1

We want to find a scalar function $f(x,y)$ such that $\nabla f(x,y)=\mathbf f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j$.
So we need to have
$\dfrac{\partial f}{\partial x}=3+4xy^2$
Integrating both sides with respect to $x$ gives
$f(x,y)=3x+2x^2y^2+g(y)$
Differentiating with respect to $y$ gives
$\dfrac{\partial f}{\partial y}=4x^2y+\dfrac{\mathrm dg}{\mathrm dy}=4x^2y$$\implies\dfrac{\mathrm dg}{\mathrm dy}=0$$\implies g(y)=C$
So we find that
$f(x,y)=3x+2x^2y^2+C$
By the fundamental theorem of calculus, we then know the line integral depends only on the values of $f(x,y)$ at the endpoints of the path. Therefore

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(3,\frac13\right)-f(1,1)=11-5=6$