Question

F=e−yi−xe−yj is conservative. find a scalar potential f and evaluate the line integral over any smooth path c connecting a(0,0) to b(1,1).

Answer 1

If $\mathbf F$ is conservative, then there is a scalar function $f$ such that

$\nabla f(x,y)=\mathbf F(x,y)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=e^{-y}\,\mathbf i-xe^{-y}\,\mathbf j$


Setting the first components equal to one another, we can integrate both sides to find

$\dfrac{\partial f}{\partial x}=e^{-y}\implies f(x,y)=xe^{-y}+g(y)$

Differentiating both sides with respect to $y$ gives

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

so that

$f(x,y)=xe^{-y}+C$

By the fundamental theorem of calculus, we have that

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(1,1)-f(0,0)=\frac1e$