I have the curve and I wish to integrate the curve: (x^2 dx + y^2 dy + z^2 dz).
Notation wise: A "C" as in to displace the bottom limit of an integral (I will attach an image so that it is easier to visualise).
Image: https://i.stack.imgur.com/5eERo.jpg
And the aforementioned curve consists of the line segments from (zero,one,zero) to (one,zero,one) and then from (one,zero,one) to (two,one,three).
I anticipate that there will be a kink/sharp twist in said curve.
With the boldened information being the most relevant - how would I solve this integral.
So far I have got this:
x=xA+t(xB−xA) dx=(xB-xA)dt.
y=yA+t(yB−yA) dy=(yB-yA)dt z=zA+t(zB−zA) dz=(zB-zA)dt.
so
$(ds)2=(dx)2+(dy)2+(dz)2$
$$ds=||\vec{AB}||dt$$
[Although I express doubt and skepticism whether this is right.]
Thank you for any help/assistance given.
Note: The image probably surmises the gist of my question a lot better.
-Nomad609
Hint for the second question.
The parametric equations of the segment $[A(0,1,0),B(1,0,1)]$ are
$x=0+t(1-0)=t$
$y=1+t(0-1)=1-t$
$z=0+t(1-0)=t$
thus
$x^2=z^2=t^2$ ,$y^2=1+t^2-2t$
$dx=-dy=dz=dt$
if we replace in the integral, we get
$$\int_0^1(t^2-(1+t^2-2t)+t^2)dt=\frac{1}{3}.$$
For the other segment, we have
$x=1+t, \; y=t, \; z=1+2t$
$dx=dy=dt, \; dz=2dt$
if we replace, we find
$$\int_0^1(10t^2+10t+3)dt=\frac{34}{3}.$$