Use the Fundamental Theorem of Line Integrals, if applicable,

to evaluate the integrals.

$\mathbf{F}(x,y)=\left(\frac{x{e}^x}{(x+y+4{)}^2},\frac{-e^x}{(x+y+4{)}^2}\right)$with $C$the cardioid given by $r=(1+2\sin \theta )$from $\theta =0$to$\theta =\mathrm{\pi }$

The given vector field is

$r=(1+2\sin \theta )$from $\theta =0$to$\theta =\mathrm{\pi }$

Assuming curve $C$is graph of vector function $r\left(t\right)$with $P=\mathbf{r}\left(a\right)$and $Q=\mathbf{r}\left(b\right)$$Q=\mathbf{r}\left(b\right)$

Let $F$is a conservative vector field with $\mathbf{F}=\nabla f$on an open, connected, and simply connected domain containing the curve $C$then,

${\int }_C\mathbf{F}·d\mathbf{r}=f\left(Q\right)-f\left(P\right)$

The vector field is

Vector field is conservative if$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$

$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$

For the given vector field and

width="151">F1(x,y)=xex(x+y+4)2

and

width="151">F2(x,y)=-ex(x+y+4)2

$\frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}\left(\frac{x{e}^x}{(x+y+4{)}^2}\right)$

$=\frac{-2x{e}^x}{(x+y+4{)}^3}$

and

$\frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\left(\frac{-e^x}{(x+y+4{)}^2}\right)$

$=\frac{-e^x(x+y+2)}{(x+y+4{)}^3}$

Hence,$\frac{\partial F_1}{\partial y}\ne \frac{\partial F_2}{\partial x}$

Vector field don't satisfy these conditions.

Therefore vector field is not conservative, hence the Fundamental Theorem of Line integral is not applied.