Computing another line integral

In summary, the homework statement is that a semi-circle on the sphere x^2+y^2+z^2 = 2 passes through the point (1,1,0). The curve C is the curve of constant \phi which passes through the point (1,1,0). At this point \sin \theta = 1 and so \cos\phi = \sin\phi = 1/\sqrt 2 on C.
  • #1
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Homework Statement



Let C be the semi-circle on the sphere [itex]x^2+y^2+z^2 = 2[/itex] from [itex]N = (0,0,\sqrt{2})[/itex] to [itex]S = (0,0, - \sqrt{2})[/itex] which passes through the point [itex](1,1,0)[/itex]

Note that x=y for all (x,y,z) on C. Evaluate the integral :

[itex]\int_C z^2dx + 2x^2dy +xydz[/itex]

Hint : Use as your parameter the angle [itex]θ[/itex] subtended at the origin by the arc NP for a point P on C.

Homework Equations



N/A

The Attempt at a Solution



So I wasn't sure how to get this one going. I'm told that C is a semi-circle on the sphere [itex]x^2+y^2+z^2 = 2[/itex] from one endpoint N to the other endpoint S which passes through (1,1,0).

So I know my first step is to parametrize using the angle θ.

So : [itex]x = cosθ, y = sinθ, z = ?[/itex] I'm thinking that z = θ. As for the interval of θ, I'm not quite sure.

Once I set the integral up, it will be easy to evaluate. I've never had a case of 3 variables over anything but lines so I'm a bit confused. I'm also thinking I may have to split this integral.

Thanks for any help in advance.
 
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  • #2
I would start with the normal parametrization of a sphere of radius [itex]\sqrt 2[/itex], which is what the hint is suggesting:
[tex]
x = \sqrt 2 \cos\phi \sin\theta \\
y = \sqrt 2 \sin\phi \sin\theta \\
z = \sqrt 2 \cos \theta
[/tex]
where [itex]0 \leq \phi < 2\pi[/itex] and [itex]0 \leq \theta \leq \pi[/itex].

The curve C is the curve of constant [itex]\phi[/itex] which passes through the point (1,1,0). At this point [itex]\sin \theta = 1[/itex] and so
[tex]
1 = \sqrt 2 \cos\phi \\
1 = \sqrt 2 \sin\phi
[/tex]
so that [itex]\cos\phi = \sin\phi = 1/\sqrt 2[/itex] on C.
 
  • #3
The hint is slightly misleading, IMO, since it is talking about ##\theta## instead of ##\phi##. If you think about the curve C in terms of spherical coordinates, ##\rho## is fixed at ##\sqrt{2}##, and the only thing that varies is ##\phi##, which ranges from 0 to ##\pi##. In spherical coordinates, ##\theta## is the angle that a vector in the x-y plane makes with the positive x-axis. ##\phi## is the angle that a vector makes with the positive z-axis.
 
  • #4
Thanks for the pointers guys, the only problem is I'm still not allowed to use spherical coordinates I believe as we have not been taught it yet.

I sat down and thought about it for a bit though. That note of x=y actually has some usage I believe.

If x = y, then [itex]2x^2 + z^2 = 2[/itex] or [itex]2y^2 + z^2 = 2[/itex]

I'm thinking that I can solve one of those to get my dx, dy and dz if I'm not mistaken?

Hopefully my professor gets to spherical and cylindrical coordinates soon.
 
  • #5
Zondrina said:
Thanks for the pointers guys, the only problem is I'm still not allowed to use spherical coordinates I believe as we have not been taught it yet.
Are you sure? This doesn't seem to me to be a valid reason for not using spherical coordinates, since they aren't really related to integration techniques.

If you really can't use spherical coordinates, the thing about the curve C is that the radius is constant and the thing that varies is the angle, as measured from the z-axis. The radius is ##\sqrt{x^2 + y^2 + z^2}##.
Zondrina said:
I sat down and thought about it for a bit though. That note of x=y actually has some usage I believe.

If x = y, then [itex]2x^2 + z^2 = 2[/itex] or [itex]2y^2 + z^2 = 2[/itex]

I'm thinking that I can solve one of those to get my dx, dy and dz if I'm not mistaken?

Hopefully my professor gets to spherical and cylindrical coordinates soon.
 
  • #6
Mark44 said:
Are you sure? This doesn't seem to me to be a valid reason for not using spherical coordinates, since they aren't really related to integration techniques.

If you really can't use spherical coordinates, the thing about the curve C is that the radius is constant and the thing that varies is the angle, as measured from the z-axis. The radius is ##\sqrt{x^2 + y^2 + z^2}##.

I asked him personally today and he told me that we could not use them which is why he gave the hint x = y.

I suppose you're now hinting that my last post was on the right track then? If so then solving and taking derivatives won't be a problem.
 
  • #7
Would help if you drew stuff you know. Also, how about just doing the first one for starters:

[tex]\int_C z^2 dx[/tex]

and keep in mind it's a half-circle along the diagonal at [itex]\pi/4[/itex] right so that's x=y. For starters, if that's so, then when we do the z^2, we'll have:

[tex]z^2\biggr|_{C}=2-(x^2+y^2)=2-2x^2[/tex]

and since you want to parameterize it in terms of [itex]\theta[/itex], we could write the integral (for starters) as:

[tex]\int_{\pi/2}^{-\pi/2} z^2dx,\quad z=z(\theta), x=x(\theta)[/tex]

Maybe that's confussing though and that direction is a little messy. So how about I just do it in the reverse direction then I could consider the integral:

[tex]\int_{\pi/2}^{-\pi/2} z^2dx=-\int_{-\pi/2}^{\pi/2} z^2dx[/tex]

But just do half of it for starters and consider:

[tex]-\int_0^{\pi/2} (2-2x^2) dx[/tex]

So what's [itex]x(\theta)[/itex] along that curve now? Keep in mind it's in 3-D so first get the diagonal distance to the point (x,x) as a function of theta, then do a cosine on it to get x right?

I think so anyway. Haven't checked this.
 
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  • #8
Wait wait... that last post confused me a bit.

So if x = y, then we have two scenarios.

[itex]2x^2+z^2 = 2[/itex] and [itex]2y^2+z^2 = 2[/itex]

If [itex]2x^2+z^2 = 2[/itex], then [itex]x = ± \sqrt{1- \frac{z^2}{2}}[/itex], similarily for y we get [itex]y = ± \sqrt{1- \frac{z^2}{2}}[/itex] since x=y.

Solving either one of them for z we get [itex]z = ± \sqrt{2} \sqrt{1-x^2}[/itex]

Now I'm a bit hesitant with what to do with these when finding my dx, dy and dz or am I completely off here? Also since I have abunch of plus minus scenarios, I'm not sure which to choose?
 
  • #9
Zondrina said:
Wait wait... that last post confused me a bit.

So if x = y, then we have two scenarios.

[itex]2x^2+z^2 = 2[/itex] and [itex]2y^2+z^2 = 2[/itex]

If [itex]2x^2+z^2 = 2[/itex], then [itex]x = ± \sqrt{1- \frac{z^2}{2}}[/itex], similarily for y we get [itex]y = ± \sqrt{1- \frac{z^2}{2}}[/itex] since x=y.

Solving either one of them for z we get [itex]z = ± \sqrt{2} \sqrt{1-x^2}[/itex]

Now I'm a bit hesitant with what to do with these when finding my dx, dy and dz or am I completely off here? Also since I have abunch of plus minus scenarios, I'm not sure which to choose?

Gotta draw stuff. If you did then you'd know for as least my integral:

[tex]\int_0^{\pi/2} (2-2x^2)dx[/tex]

everything is positive since I'm integrating over the x-y plane in the first quadrant over the sphere surface over that quadrant which is also positive z.

Take a point z on that contour between 0 and pi/2 over the x-y plane. Drop a line down to the x-y plane to the point (x,y). A diagonal line from the origin to the point (x,x) say is [itex]\rho[/itex]. Then we can write:

[tex]\rho^2=2x^2[/tex]

Now [itex]\rho[/itex] become the base of another triangle, the points are from (0,0,0) to (x,x,0) to (x,x,z). We know what the hypotnuse is, [itex]\sqrt{2}[/itex] and the angle between the x-y plane and the hypotnuse is say our [itex]\theta[/itex]. Then:

[tex]\cos(\theta)=\frac{\rho}{\sqrt{2}}[/tex]

or [tex]\rho=\sqrt{2}\cos(\theta)=\sqrt{2}x[/tex]

Then:

[tex]x(\theta)=\cos(\theta)[/tex]

Ok, that's the first one. Try and do the other two now if I've succeeded in helping you understand the principle.
 
  • #10
Mark44 said:
In spherical coordinates, ##\theta## is the angle that a vector in the x-y plane makes with the positive x-axis. ##\phi## is the angle that a vector makes with the positive z-axis.

There are two exactly opposite conventions as to what [itex]\phi[/itex] and [itex]\theta[/itex] mean.
 
  • #11
pasmith said:
There are two exactly opposite conventions as to what [itex]\phi[/itex] and [itex]\theta[/itex] mean.
That may be, but the calculus texts that I remember teaching from used the convention that I described.
 
  • #12
Mark44 said:
That may be, but the calculus texts that I remember teaching from used the convention that I described.

In physics it's almost always the other way around. Not that this looks like a physics problem.
 

FAQ: Computing another line integral

1. What is a line integral in computing?

A line integral is a mathematical concept that involves integrating a function over a curve or a line. It is used in computing to calculate the work done by a vector field along a given path.

2. How is a line integral computed?

To compute a line integral, you first need to parametrize the curve or line and then integrate the function along the parametrized curve. This involves taking the dot product of the function with the tangent vector of the curve and integrating it with respect to the parameter.

3. What are the applications of computing a line integral?

Line integrals have various applications in physics, engineering, and other fields. They are used to calculate work, mass, and energy in vector fields. They can also be used to find the length of a curve, the center of mass of an object, and the flux of a vector field.

4. What is the difference between a line integral and a double integral?

A line integral is calculated along a curve or a line, while a double integral is calculated over a two-dimensional region. Line integrals involve a single variable parameter, while double integrals involve two variables.

5. Can line integrals be generalized to higher dimensions?

Yes, line integrals can be extended to higher dimensions. In three dimensions, they are known as surface integrals, and in higher dimensions, they are called volume integrals. These integrals are used to calculate quantities such as flux, surface area, and volume in vector fields.

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