# Computing another line integral

1. Jan 14, 2013

### Zondrina

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

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

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

$\int_C z^2dx + 2x^2dy +xydz$

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

2. Relevant equations

N/A

3. 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 $x^2+y^2+z^2 = 2$ 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 : $x = cosθ, y = sinθ, z = ?$ 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.

2. Jan 14, 2013

### pasmith

I would start with the normal parametrization of a sphere of radius $\sqrt 2$, which is what the hint is suggesting:
$$x = \sqrt 2 \cos\phi \sin\theta \\ y = \sqrt 2 \sin\phi \sin\theta \\ z = \sqrt 2 \cos \theta$$
where $0 \leq \phi < 2\pi$ and $0 \leq \theta \leq \pi$.

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
$$1 = \sqrt 2 \cos\phi \\ 1 = \sqrt 2 \sin\phi$$
so that $\cos\phi = \sin\phi = 1/\sqrt 2$ on C.

3. Jan 14, 2013

### Staff: Mentor

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. Jan 14, 2013

### Zondrina

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 $2x^2 + z^2 = 2$ or $2y^2 + z^2 = 2$

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. Jan 14, 2013

### Staff: Mentor

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 $\sqrt{x^2 + y^2 + z^2}$.

6. Jan 14, 2013

### Zondrina

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 wont be a problem.

7. Jan 15, 2013

### jackmell

Would help if you drew stuff you know. Also, how about just doing the first one for starters:

$$\int_C z^2 dx$$

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

$$z^2\biggr|_{C}=2-(x^2+y^2)=2-2x^2$$

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

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

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:

$$\int_{\pi/2}^{-\pi/2} z^2dx=-\int_{-\pi/2}^{\pi/2} z^2dx$$

But just do half of it for starters and consider:

$$-\int_0^{\pi/2} (2-2x^2) dx$$

So what's $x(\theta)$ 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.

Last edited: Jan 15, 2013
8. Jan 15, 2013

### Zondrina

Wait wait... that last post confused me a bit.

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

$2x^2+z^2 = 2$ and $2y^2+z^2 = 2$

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

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

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. Jan 15, 2013

### jackmell

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

$$\int_0^{\pi/2} (2-2x^2)dx$$

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 $\rho$. Then we can write:

$$\rho^2=2x^2$$

Now $\rho$ 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, $\sqrt{2}$ and the angle between the x-y plane and the hypotnuse is say our $\theta$. Then:

$$\cos(\theta)=\frac{\rho}{\sqrt{2}}$$

or $$\rho=\sqrt{2}\cos(\theta)=\sqrt{2}x$$

Then:

$$x(\theta)=\cos(\theta)$$

Ok, that's the first one. Try and do the other two now if I've succeeded in helping you understand the principle.

10. Jan 16, 2013

### pasmith

There are two exactly opposite conventions as to what $\phi$ and $\theta$ mean.

11. Jan 16, 2013

### Staff: Mentor

That may be, but the calculus texts that I remember teaching from used the convention that I described.

12. Jan 16, 2013

### Dick

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