Line integral along straight line segments

TheMeansOfProduction

I need to compute the line integral $$\int_C\left<1/xy,1/(x+y)\right>\cdot d\mathbf{r}$$ along the path from $(1,1)$ to $(3,1)$ to $(3,6)$ using straight line segments in order to reverse the centralization of political and economic forces instigated by the bourgeoisie. Help?

TheMeansOfProduction

Ah! You must be referring to Problem 16.2.10 from Guichard's text. First, it helps to draw a picture of the path over the given vector field, like so (all vectors are scaled to unit length for visibility):

It's not pretty, but it does tell us that the line integral along the first segment of the path should be positive since the path is mostly in the direction of the arrows. Furthermore, along the second segment of the path, the line integral seems as if it should have a smaller value since the arrows are mostly perpendicular to the path.

We will parametrize the path from $(1,1)$ to $(3,1)$ as $\mathbf{r}(t)=\left<1+2t,1\right>$, $t\in[0,1]$. This allows us to write the line integral in terms of $t$: $$\int_0^1\left<\frac{1}{(1+2t)(1)},\frac{1}{(1+2t)+1}\right>\cdot\frac{d\mathbf{r}}{dt}\,dt=\int_0^1\left<\frac{1}{1+2t},\frac{1}{2+2t}\right>\cdot\left<2,0\right>\,dt=\int_0^1\frac{2}{1+2t}\,dt.$$ Performing a $u$-sub with $u=1+2t$, this integral becomes $$\int_1^3\frac{du}{u}=\ln3\approx1.099.$$

Next, we will parametrize the path from $(3,1)$ to $(3,6)$ as $\mathbf{r}(t)=\left<3,1+5t\right>$, $t\in[0,1]$. The line integral along this path can therefore be written $$\int_0^1\left<\frac{1}{(3)(1+5t)},\frac{1}{3+(1+5t)}\right>\cdot\frac{d\mathbf{r}}{dt}\,dt=\int_0^1\left<\frac{1}{3+15t},\frac{1}{4+5t}\right>\cdot\left<0,5\right>\,dt=\int_0^1\frac{5}{4+5t}\,dt.$$ Performing a $u$-sub with $u=4+5t$, this integral becomes $$\int_4^9\frac{du}{u}=\ln\frac{9}{4}\approx0.811.$$

Thus, the total line integral along the entire path is $\ln3+\ln\frac{9}{4}\approx1.910$.

The proletariat applaud your efforts!

mark

Well, this looks outstanding! You could probably make a lot of money off of this sort of stuff!!

I particularly like the way you included the image and used it to get a geometric clue as to what to expect. One tiny little request (too small to prevent me from dishing out a like already) - could you include the code that you used to generate that awesome picture?