Answer
$-1$
Work Step by Step
Vector field $F(x,y) = $ $(1+xy)e^{xy}i$ + $x^{2}e^{xy}j$
Since $F(x,y)$ is assumed conservative, we have
$f_{x} = (1+xy)e^{xy}$
$f_{y} = x^{2}e^{xy}$
Thus, integrating $f_{x}$ with respect to x, we get
$f = xe^{xy} + g(y)$ [g(y) : is a function of y]
Now taking the derivative of $f$ with respect to y, we get
$f_{y} = x^{2}e^{xy} + g'(y)$
Since $f_{y} = x^{2}e^{xy}$ and $f_{y} = x^{2}e^{xy} + g'(y)$, we now get
$g'(y) = 0$
Therefore,
$f = xe^{xy}$
Using the fundamental theorem of line integrals
$\int_C \nabla f \cdot dr = f(r(b)) - f(r(a)) $ ,
The curve $C$ is defined by the parameter $r(t) = \cos(t)i + 2\sin(t)j$, $0â¤tâ¤\pi/2$
Thus, we get
$r(b) = r(\pi/2)$
$r(a) = r(0)$
Hence, $r(b) = (0,2)$ and $r(a) = (1,0)$, and we can then get
$f(0,2) - f(1,0) = 0 - 1 = -1$
