91Ó°ÊÓ

The given vector field is

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

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 $\mathbf{F}=∇f$on an open, connected, and simply connected domain containing the curve $C$then,

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

The vector field is

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

$\frac{∂F_1}{∂y}=\frac{∂F_2}{∂x}$

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

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

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

and

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

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

Hence,$\frac{∂F_1}{∂y}≠\frac{∂F_2}{∂x}$

Vector field don't satisfy these conditions.

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