Calculating the Line Integral for the Intersection of Two Surfaces

[santais]

Homework Statement

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

x²+z²=1

and

y = x²

Calculate

[itex]\int_c{\sqrt{1+4x^2 z^2}}[/itex]

Homework Equations

So basicly I'm completely lost on this assignment. So far I can see, that [itex]x^2+z^2=1[/itex] describes a circle in x-z plane. And [itex]y = x^2[/itex] 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.

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:)

 

[HallsofIvy]

First, no, [itex]x^2+ z^2= 1[/itex] 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, [itex]y= x^2[/itex] 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 [itex]x^2+ z^2= 1[/itex] and [itex]y= x^2[/itex], we can immediately see that [itex]y+ z^2= 1[/itex] of [itex]y= 1- z^2[/itex].

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

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

 

[DryRun]

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

 

[santais]

First, no, [itex]x^2+ z^2= 1[/itex] 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, [itex]y= x^2[/itex] 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 [itex]x^2+ z^2= 1[/itex] and [itex]y= x^2[/itex], we can immediately see that [itex]y+ z^2= 1[/itex] of [itex]y= 1- z^2[/itex].

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

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

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.

 

[HallsofIvy]

Yes, that's correct.

 

[DryRun]

Well, I've been working on this problem, and it's not as simple as it looks:
[tex]ds=\sqrt{2-4\sin \theta \cos \theta + 4 \sin ^2 \theta \cos ^2 \theta + \cos ^2 \theta}\,.d\theta[/tex]
So, the line integral becomes:
[tex]\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[/tex]
Is that correct?