Computing another line integral

[STEMucator]

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.

 

[pasmith]

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.

 

[Mark44]

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.

 

[STEMucator]

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.

 

[Mark44]

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}##.

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.

 

[STEMucator]

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.

 

[jackmell]

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.

 

[STEMucator]

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?

 

[jackmell]

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.

 

[pasmith]

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.

 

[Mark44]

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.

 

[Dick]

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.