Question

G find the value of the line integral. f · dr c (hint: if f is conservative, the integration may be easier on an alternative path.) f(x,y) = yexyi + xexyj

Answer 1

Assuming the field is given by


`\mathbf f(x,y)=ye^(xy)\,\mathbf i+xe^(xy)\,\mathbf j`

then because
`\mathbf f(x,y)` is continuous, there is some scalar potential function
`f(x,y)` such that
`\\abla f(x,y)=\mathbf f(x,y)`, i.e. the vector field is conservative. So


`(\partial f(x,y))/(\partial x)=ye^(xy)`

`\displaystyle\int(\partial f)/(\partial x)=\frac{ye^(xy)}y+g(y)=e^(xy)+g(y)`


`(\partial f(x,y))/(\partial y)=(\partial(e^(xy)+g(y)))/(\partial y)`

`xe^(xy)=xe^(xy)+g'(y)`

`0=g'(y)`

`\implies g(y)=C`


`\implies f(x,y)=e^(xy)+C`

The value of the line integral then depends on only the endpoints of the path and the gradient theorem (fundamental theorem of calculus for line integrals) applies.