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 $\nabla f(x,y)=\mathbf f(x,y)$, i.e. the vector field is conservative. So

$\dfrac{\partial f(x,y)}{\partial x}=ye^{xy}$
$\displaystyle\int\dfrac{\partial f}{\partial x}=\dfrac{ye^{xy}}y+g(y)=e^{xy}+g(y)$

$\dfrac{\partial f(x,y)}{\partial y}=\dfrac{\partial(e^{xy}+g(y))}{\partial y}$
$xe^{xy}=xe^{xy}+g'(y)$
$0=g'(y)$
$\implies g(y)=C$

$\implies f(x,y)=e^{xy}+C$

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.