Compute the work done by the vector field $$F(x,y)=\left(y, xy-\frac{9y^2}{x}\right),$$ when a particle is moving along the curve $x^2/9-y^2=1,$ from $(3\sqrt{10},3)$ to $(3\sqrt{5},-2).$
The curve goes in negative direction $-\gamma$. I have that $P(x,y)=y, \ Q(x,y)=xy-\frac{9y^2}{x}$ so
$$W=\int_{-\gamma}Pdx+Qdy = -\int_{\gamma}Pdx+Qdy...$$
However, I first need to parameterize my integrand and I'm not sure what I should set $x$ and $y$ to. Is there any systematic method to determine an appropriate parameterization? If it's a circle it's easy by using $x=\cos(t) \ y=\sin(t).$
I'm going to make an answer to avoid talking in the comments, and it would also be too long for a comment.
From @Salahamam_ Fatima's idea, we can use the following parametrization: $$(x,\,y)=(3\cosh(t),\,\sinh(t))$$ Your integral is: $$\int\vec{F}\cdot\mathrm{d}\vec{r}=\int \vec{F}\cdot\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\mathrm{d}t$$ $$=\int\left(\sinh(t),\,3\sinh(t)\cosh(t)-3\frac{\sinh^2(t)}{\cosh(t)}\right)\cdot(3\sinh(t),\,\cosh(t)) \mathrm{d}t$$ $$=\int\left[3\sinh^2(t)+3\sinh(t)\cosh^2(t)-3\sinh^2(t)\right]\mathrm{d}t$$ $$=\int\left[3\sinh(t)\cosh^2(t)\right]\mathrm{d}t$$ $$=3\int\left[\sinh(t)\cosh^2(t)\right]\mathrm{d}t$$ $$=3\left.\frac{\cosh^3(t)}{3}\right|_{\sinh^{-1}(3)}^{\sinh^{-1}(-2)}=\left.\cosh^3(t)\right|_{\sinh^{-1}(3)}^{\sinh^{-1}(-2)}=5\sqrt{5}-10\sqrt{10}$$