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


`\\abla f(x,y)=\mathbf F(x,y)\iff(\partial f)/(\partial x)\,\mathbf i+(\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


`(\partial f)/(\partial x)=e^(-y)\implies f(x,y)=xe^(-y)+g(y)`

Differentiating both sides with respect to
`y` gives


`(\partial f)/(\partial y)=-xe^(-y)+(\mathrm dg)/(\mathrm dy)=-xe^(-y)`

`\implies(\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)\\abla f\cdot\mathrm d\mathbf r=f(1,1)-f(0,0)=\frac1e`