Homework Help: Calculate the line integral

1. Apr 23, 2012

santais

1. The problem statement, all variables and given/known data

Observe the curve C, which is given as the intersection between the surfaces:

x2+z2=1

and

y = x2

Calculate

$\int_c{\sqrt{1+4x^2 z^2}}$

2. Relevant equations

So basicly I'm completely lost on this assignment. So far I can see, that $x^2+z^2=1$ describes a circle in x-z plane. And $y = x^2$ is just a parabolic in the y plane.

So my guess that the limits would go from 0 >= θ >= 2Pi and 0 >= r >= 1, once it has been rewritten in polar coordinates.

3. The attempt at a solution

So far I havn't been able to get going. The thing is, that makes this troublesome, is that the circle in the x-z plane opposite the x-y plane, which it usually is in, in an assignment.

Hope there is some of you out there, that could give me a hint to get going:)

2. Apr 23, 2012

HallsofIvy

First, no, $x^2+ z^2= 1$ does NOT describe a circle in the x-z plane. That would be if y= 0 but since there is no y in the equation, y can be anything. The equation describes a circular cylinder with axis along the y axis. Similarly, $y= x^2$ is a parabolic cylinder extending along the z-direction. In general, a single equation in three dimensions describes a surface, not a curve.

Of course, the intersection of those two surfaces is a curve. With $x^2+ z^2= 1$ and $y= x^2$, we can immediately see that $y+ z^2= 1$ of $y= 1- z^2$.

So we could use z itself as parameter though in that case we would have to do it as two separate integrals:
1) $x= \sqrt{1- t^2}$, $y= 1- t^2$, $z= t$ and
2) $x= -\sqrt{1- t^2}$, $y= 1- t^2$, $z= t$

It might be better to use the "standard" circle parameterization:
$x= cos(t)$, $y= 1- sin^2(t)$, $z= sin(t)$.

3. Apr 23, 2012

sharks

Here's an extension from HallsofIvy's advice:
$$\int_C{\sqrt{1+4x^2 z^2}}\,.ds$$ where C is the curve $y= 1- z^2$
Now, to find $ds$:
$$ds= \sqrt { (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}\,.dt$$

4. Apr 23, 2012

santais

Thanks for the quick reply :)

It sure did lighten it up a lot. I see that i have misunderstood what geometric figure the functions describe. What I meant by a circle in the x-z plane was, that the projection of the function to the x-z plane, would give show a circle.

Anyway so now it's just rewriting it to "standard" circle parameterization, aka Polar coordinates? And calculate the rest of the integral frmo there.

5. Apr 23, 2012

HallsofIvy

Yes, that's correct.

6. Apr 23, 2012

sharks

Well, i've been working on this problem, and it's not as simple as it looks:
$$ds=\sqrt{2-4\sin \theta \cos \theta + 4 \sin ^2 \theta \cos ^2 \theta + \cos ^2 \theta}\,.d\theta$$
So, the line integral becomes:
$$\int^{\theta = 2\pi}_{\theta = 0} \sqrt{1+ 4 \sin ^2 \theta \cos ^2 \theta}.\sqrt{2-4\sin \theta \cos \theta + 4 \sin ^2 \theta \cos ^2 \theta + \cos ^2 \theta}\,.d\theta$$
Is that correct?