I would appreciate if someone could help me with this question. I need to find an arc length of a space curve which is an intersection of two surfaces $4ax=(y+z)^2$ and $4x^2+3y^2=3z^2$, from the origin to the point $(x,y,z)$.
I am trying to parametrize the sought for curve with $t$ or to express two of the variables with the third one so I can then calculate the derivatives and plug them into the formula for arc length differential and then integrate it.
But, I am stuck at the very beginning of the process. Thanks for any hints regarding the parametrization of this specific example.
Thanks to Sammy Black's hint I moved forward with the problem. This is what I got:
Using the substitution: $u=z+y$ and $v=z-y$ I get:
$x(u)=\frac{u^2}{4a}$ and $v(u)=\frac{u^3}{12a^2}$.
After differentiation: $x'(u)=\frac{u}{2a}$ and $v'(u)=\frac{u^2}{4a^2}$.
Now I use the formula $ds=\sqrt{(1+(x'(u))^2+(v'(u))^2)}du$ to get $ds=\frac{1}{4a^2}\sqrt{(u^2+2a^2)^2+12a^4}du$.
The limits of integration will be $u=0$ and $u=z+y$ which finally gives me $arc length=\int_0^{z+y}\frac{1}{4a^2}\sqrt{(u^2+2a^2)^2+12a^4}du$ and I really don't know how ot crack this integral.
Can someone please help me with that integral, or see if I made a mistake somewhere on the way.
