13.4#5

Hello Professor,
You asked me to email you about this question so you could explain it on the blog.
Much Appreciated,


*********************************

Ah. This is a different problem than the one I thought we were talking about.   Notice that the problem says that the path is "nonclosed"?  This means that you don't have a closed loop, just the three line segments AB+BC+CD, but you don't have DA also, so Green's theorem doesn't directly apply.  You could solve this problem directly just by integrating F=<sin(x)+3y, 2x+y> and integrating
∫ F.dr_1 + ∫ F.dr_2 + ∫ F.dr_3 where r_1(t)=t*<2,2> for 0≤t≤1, r_2(t)=(1-t)*<2,2> + t*<2,4> for 0≤t≤1, r_3(t)=(1-t)*<2,4> + t*<0,6> for 0≤t≤1. Or you *could* still use Green's theorem if you're willing to be a little sneaky: since ∂F_2/∂x = 2 and ∂F_1/∂y = 3, Green's theorem implies that
∬_D (-1)dA - ∫ F.dr_4 = ∫ F.dr_1 + ∫ F.dr_2 + ∫ F.dr_3 ,  where D is the enclosed region and where r_4(t)=(1-t)*<0,6> . This is a little bit easier approach since ∬_D (-1)dA = -Area = -8, and
r'_4(t)= -<0,6> so ∫ F.dr_4 = ∫_0^1<18(1-t), 6(1-t)>.<0, -6> dt = -36 ∫_0^1(1-t) dt.