I'm having some trouble solving a couple of problems:
- I know this one must be pretty easy but can't find the way to solve it. I need to find the arc length of a curve described by $ r=1- \theta ; 1\leq \theta \leq 2.$
From my notes, this should be solved with \int_{C} f(\sigma (t)) \left \|\sigma '(t)\right \| dt I would use $\sigma (t)= (1-t, t)$ since that describes $C$, but what $F$ am I supposed to use?
- Again, I feel like this should be really easy, but can't figure it out:
$C = {(x,y,z,): y = 1 - x^2 ; x + y + z = 1 ; x,y \geq 0} $
a) I need to find a regular parametrization of $C$ that starts in $(0,1,0)$ and ends in $(1,0,0).$
b) I need to find $ \int_{C} F. ds$ with $ F = (2x, y, -x).$
My problem here is finding the parametrization. I think I could use $ 1 \leq z \leq x + y - 1 , 0 \leq x \leq a(x,y,z) , 0 \leq y \leq b(x,y,z).$
I can't really decide what those $a$ and $b$ should be. Once that is done, I think solving b) should be pretty easy.
Anyway, I'll be grateful for any pointers on how to solve this and similar problems, since I can't seem to grasp the concepts behind most of this.
EDIT 1: Regarding problem 1, I think I should do $\int_{1}^{2} \left \|\frac{\partial (1-\theta ; \theta )}{\partial \theta } \right \| d\theta = \sqrt{2} $ and that should be the length. I had actually tried that but didn't seem right, but after looking around a little bit more and finding nothing, I think this is it.
EDIT 2: EDIT 1 was wrong. I had to use $\int_{a}^{b} \sqrt{(r(\theta))^{2} + (\frac{\partial{r(\theta)}}{\partial{\theta}})^{2}}d\theta$ with $r(\theta)=1-\theta$ which gives me $\int_{1}^{2} \sqrt{2-2\theta+(\theta)^{2}} d\theta $ which is pretty hard to solve. I'm guessing I still have something wrong.
Regarding problem 2: I wrote $\sigma(t) = (t, 1-t^{2}, t^{2}-t) \in C^1$ and also \sigma'(t) \neq (0,0,0) , noting that $\sigma(0)=(0,1,0)$ and $\sigma(1)=(1,0,0).$
So point a) is done. Now I needed to solve \int_{C} F. ds \Rightarrow \int_{0}^{1} F(\sigma(t))\sigma'(t)dt I got $\int_{0}^{1} (2t,2-2t^2, -t)(1,-2t, 2t-1)dt = \frac{-1}{6}$ This one I think is right, but I'm not sure what the result means. Does F goes through $\sigma$ the other way around?
EDIT 3: The integral from the second excercise should be $\int_{0}^{1} (2t,1-t^2, -t)(1,-t,2t-1) dt = \frac{1}{3}$ which is a lot nicer. Thanks J.M.
Still not sure about the first one though.