Question

Consider f and c below. f(x, y) = x2y3i + x3y2j,

Answer 1

If there is a scalar function $f(x,y)$ such that $\mathbf f(x,y)=\nabla f(x,y)$, then it would satisfy

$\dfrac{\partial f}{\partial x}=x^2y^3$
$\dfrac{\partial f}{\partial y}=x^3y^2$

From the first PDE, we have

$f(x,y)=\dfrac13x^3y^3+g(y)$

Take the derivative with respect to $y$ to get

$\dfrac{\partial f}{\partial y}=x^3y^2+\dfrac{\mathrm dg}{\mathrm dy}=x^3y^2$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

So,

$f(x,y)=\dfrac13x^3y^3+C$


By the fundamental theorem of calculus,

$\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f(x,y)\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))$
$=f(-1,3)-f(0,0)=-9$