Find F.dr along C if F(x,y) = <x^2+2y^3-4x+3, -2x^3> and

[MarkSA]

Hi,

Find F.dr along C if F(x,y) = <x[sup:1bsdqrtc]2[/sup:1bsdqrtc] + 2y[sup:1bsdqrtc]3[/sup:1bsdqrtc] - 4x + 3, -2x[sup:1bsdqrtc]3[/sup:1bsdqrtc]> and C is the curve in the xy-plane starting at point (2,0) and follows the semi-circle y=sqrt(4 - x[sup:1bsdqrtc]2[/sup:1bsdqrtc]) to the point (-2,0) then followed the line segment from (-2,0) back to (2,0).

I'm not real sure where to begin with this. I thought I could break it up into two parts, using green's theorem for the semicircle and probably the parameterization for a line segment r(t) = (1-t)r[sub:1bsdqrtc]0[/sub:1bsdqrtc] + tr[sub:1bsdqrtc]1[/sub:1bsdqrtc] and integrate F(r(t)) . r'(t)dt, then add the two together.

But i'm not sure if the semicircle can really be called closed for green's theorem to work. Am I on the right track here?

 

[MarkSA]

*bump* anyone know how to work this problem? It's a type that I really need to know how to do before I take the final.

 

[galactus]

Looks like we can use Green's theorem.

\(\displaystyle \int\int_{R}\left(\frac{{\partial}g}{{\partial}x}-\frac{{\partial}f}{{\partial}y}\right)dA\)

\(\displaystyle \int\int\left(\frac{{\partial}}{{\partial}x}(-2x^{3})-\frac{{\partial}}{{\partial}y}(x^{2}+2y^{3}-4x+3)\right)dA\)

This gives us:

\(\displaystyle -6\int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\left[(x^{2}+y^{2})\right]dydx\)

We could also use polar coordinates.

\(\displaystyle -6\int_{0}^{\pi}\int_{0}^{2}(r^{3})drd{\theta}\)

and get the same.

 

[MarkSA]

Thanks!