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Integrate r = $\sqrt{x^2+y^2}$ from (0,0) to (1,1) along the path (0,0) => (1,0) => (1,1)

My professor tells me to let $dr = dxi +dyj$ where $i $ and $j$ are the standard unit vectors. I don't really see how this is possible with a scalar function. I have been parametrizing:

$$\int_Cr(x,y)dr = \int_{t_0}^tr(x_{(t)},y_{(t)})\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt$$

and i get $C_1$ (0,0) => (1,0)
$$x = t;dx/dt =1;y =0$$

But my main confusion is when I try to parameterize C2: (1,0)=>(1,1) I get an integral which I cannot solve by any conventional analytical methods: x = 1, y = t, dy/dt = 1 $$\int\sqrt{1 + t^2}dt$$

So I have surely done something wrong, any help would be greatly appreciated.

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up vote 1 down vote accepted

This suggestion $dr=dx\cdot i+dy\cdot j$ maybe a bit confusing, but perhaps better using that than this one: $dr=\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt$, which is not correct at all: in this notation $dr/dt$ gives the derivative of $r$ by $t$. Also, it seems you want to denote 2 things by $r$: the function itself and the variable point on the given path.

But why don't you just integrate $r$ on the given path? First $y=0$ is fix and $x$ goes from $0$ to $1$, then $x=1$ is fix and $y$ goes from $0$ to $1$. Two one variable integrals (and the second one is exactly that you wrote up:) $$\int\sqrt{1+y^2}dy$$ For this: have you heard about $\cosh^2-\sinh^2=1$?

So, try the following substitution: $y=\sinh v$, then $dy/dv=\cosh v$, alias, formally, $dy=\cosh v\cdot dv$. You will also need the identity for $\cosh(2v)$.

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okay thanks a lot – Cactus BAMF Oct 7 '12 at 3:51

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