Integrating $$\frac{ds}{(2y^2+1)^{3/2}}$$ on a Parabolic Curve

[sigh1342]

Homework Statement

Find $$\int_{C} \frac{ds}{(2y^2+1)^{3/2}}$$
where $$C$$ is the parabola $$ z^2=x^2+y^2 , x+z=1$$

Homework Equations

The Attempt at a Solution

I tried to parametrize the C , s.t$$ x=t, z=1-t, y=\sqrt{2t-1}$$ ,
but it seems to become a mess, and I don't know the bound of t,
and I tried to let$$ x=sin^2ω , z=cos^2ω , y = \sqrt{cos^2ω-sin^2ω}$$
Is $$ω $$from 0 to $$2\pi$$? , but it seems to not easy to compute it.
So I want to know is there any nice method for that Q.?

 

[jackmell]

Homework Statement

Find $$\int_{C} \frac{ds}{(2y^2+1)^{3/2}}$$
where $$C$$ is the parabola $$ z^2=x^2+y^2 , x+z=1$$

Homework Equations

The Attempt at a Solution

I tried to parametrize the C , s.t$$ x=t, z=1-t, y=\sqrt{2t-1}$$ ,
but it seems to become a mess, and I don't know the bound of t,
and I tried to let$$ x=sin^2ω , z=cos^2ω , y = \sqrt{cos^2ω-sin^2ω}$$
Is $$ω $$from 0 to $$2\pi$$? , but it seems to not easy to compute it.
So I want to know is there any nice method for that Q.?

You got:

$$z^2=x^2+y^2$$
$$z=1-x$$
or:
$$x=1/2-y^2/2$$
so we can parameterize that in terms of y as:

$$x(y)=1/2-y^2/2$$
$$y(y)=y$$
$$z=1-(1/2-y^2/2$$
or to make it look nicer:

$$x(t)=1/2-t^2/2$$
$$y(t)=t$$
$$z(t)=1/2+t^2/2$$
and that last parametrization will track a parabolic curve on the surface of $z^2=x^2+y^2$ for how long? Well from one end of the parabola to the other end right? How long is that if I didn't give any end point? How about -100 to 100? Further? How about -1000 to 1000? Further.

 

[sigh1342]

I got it , thank you very much :D