Say I wanted to integrate a curve f(x) such that f(x) = $x$$^2$ + $y$$^2$ +$z$$^2$ so that the integral of said curve is from this $x$$^2$ $dx$ + $y$$^2$ $dy$ + $z$$^2$ $dz$.
In looking for help, I found this from Yahoo Answers: https://answers.yahoo.com/question/index?qid=20111211091146AAl6uVn although my question is similar in some aspects I am confused how that integral is done with respect to "ds" as opposed to what mine is from the second line of this post. Is it the same thing or totally different?
And if aforementioned curve relating to my question was composed of line segments from say: $(0,1,0)$ going to $(1,0,1)$ can anyone advise me in how to start it off?
Finally should the parameterization be provided - or is this something that I will have to figure out myself.
Thank you for your help in advance, I apologise for asking several questions - I am relatively new to this forum.
We have
$x=t, y=cos(2t), z=sin(2t)$
so
$x^2+y^2+z^2=t^2+1$
and
$dx=dt, dy=-2sin(2t)dt$
$ dz=2cos(2t)dt$
which gives
$$ds=\sqrt{dx^2+dy^2+dz^2}=$$
$$dt\sqrt{(1+4)}=\sqrt{5}dt$$
thus , the integral becomes
$$I=\sqrt{5}\int_0^{2\pi}(t^2+1)dt$$
$$=\sqrt{5}(\frac{t^3}{3}|_0^{2\pi}+2\pi)$$
$$=2\pi\sqrt{5}(\frac{4}{3}\pi^2+1)$$
Now for line AB.
we have $A(0,1,0)$ and $B(1,0,1)$.
the parametric equations of line $AB$ are
$x=x_A+t(x_B-x_A)=t$ and $dx=dt$.
$y=y_A+t(y_B-y_A)=1-t$ and $dy=-dt$
$z=z_A+t(z_B-z_A)=t$ and $dz=dt$.
so
$(ds)^2=(dx)^2+(dy)^2+(dz)^2=$
$3(dt)^2$ and $ ds=\sqrt{3}dt$.
the integral is then
$$\sqrt{3}\int_0^1(1+t^2)dt=$$
$$\frac{4\sqrt{3}}{3}.$$