Question

Suppose \nabla f (x,y) = 3 y \sin(xy) \vec{i} + 3 x \sin(xy)\vec{j}, \vec{f} = \nabla f(x,y), and c is the segment of the parabola y = 3 x^2 from the point (1,3) to (4,48). then

Answer 1

I'll assume you're supposed to compute the line integral of $\nabla f$ over the given path $C$. By the fundamental theorem of calculus,

$\displaystyle\int_C\nabla f(x,y)\cdot\mathrm d\vec r=f(4,48)-f(1,3)$

so evaluating the integral is as simple as evaluting $f$ at the endpoints of $C$. But first we need to determine $f$ given its gradient.

We have

$\dfrac{\partial f}{\partial x}=3y\sin(xy)\implies f(x,y)=-3\cos(xy)+g(y)$

Differentiating with respect to $y$ gives

$\dfrac{\partial f}{\partial y}=3x\sin(xy)=3x\sin(xy)+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

and we end up with

$f(x,y)=-3\cos(xy)+C$

for some constant $C$. Then the value of the line integral is $-3\cos192+3\cos3$.