# Need some help with a line integral problem...

• Apr 25th 2010, 08:54 AM
Redding1234
Need some help with a line integral problem...
Problem: Let C be the curve of intersection of the upper hemisphere $\displaystyle x^2 + y^2 + z^2 = 4, z\geq0$ and the cylinder $\displaystyle x^2 + y^2 = 2x$, oriented counterclockwise as viewed from high above the xy-plane. Evaluate $\displaystyle \int_{C}{ydx + zdy + xdz}$.

From what I understand, we need a parametrization in terms of one variable. I found an intersection curve parametrization as:

$\displaystyle x = 2 - \frac{t^2}{2}$

$\displaystyle y = \frac{t\sqrt{4 - t^2}}{2}$

$\displaystyle z = t$

And now I'm not sure how to proceed. I suspect I will need to evaluate this integral:

$\displaystyle \int_{t=a}^b {[\frac{t\sqrt{4 - t^2}}{2} * -t} +$ $\displaystyle t * (\frac{\sqrt{4 - t^2}}{2} - \frac{t^2}{2\sqrt{4 - t^2}})+ 2 - \frac{t^2}{2}] dt$

Assuming all of that is correct, I suppose my primary question is how to find the limits of integration.
• Apr 25th 2010, 11:03 AM
Jester
Quote:

Originally Posted by Redding1234
Problem: Let C be the curve of intersection of the upper hemisphere $\displaystyle x^2 + y^2 + z^2 = 4, z\geq0$ and the cylinder $\displaystyle x^2 + y^2 = 2x$, oriented counterclockwise as viewed from high above the xy-plane. Evaluate $\displaystyle \int_{C}{ydx + zdy + xdz}$.

From what I understand, we need a parametrization in terms of one variable. I found an intersection curve parametrization as:

$\displaystyle x = 2 - \frac{t^2}{2}$

$\displaystyle y = \frac{t\sqrt{4 - t^2}}{2}$

$\displaystyle z = t$

And now I'm not sure how to proceed. I suspect I will need to evaluate this integral:

$\displaystyle \int_{t=a}^b {[\frac{t\sqrt{4 - t^2}}{2} * -t} +$ $\displaystyle t * (\frac{t\sqrt{4 - t^2}}{2} - \frac{t^2}{2\sqrt{4 - t^2}})+ 2 - \frac{t^2}{2}] dt$

Assuming all of that is correct, I suppose my primary question is how to find the limits of integration.

First, you only have half the curve. When solving for $\displaystyle y$ there's a $\displaystyle \pm$ to consider. Since your primary variable is $\displaystyle z = t$, then $\displaystyle t$ is between $\displaystyle 0$ and $\displaystyle 2$ but be careful on which way you go (remember, you gotta transverse CCW around the curve).

I think the parameterization

$\displaystyle x = 1 - \cos \theta, y = \sin \theta, \theta = 0 \to 2 \pi$

is better and much more natural.

Note: Solving for $\displaystyle z$ gives $\displaystyle z = 2 \sin \frac{\theta}{2}$.
• Apr 25th 2010, 11:55 AM
Redding1234
For the way I set it up, how could I figure out if $\displaystyle t$ is going from 0 to 2 or from 2 to 0? Is a picture necessary? I was having trouble drawing and labeling a picture of the curve.

Quote:

Originally Posted by Danny
I think the parameterization

$\displaystyle x = 1 - \cos \theta, y = \sin \theta, \theta = 0 \to 2 \pi$

is better and much more natural.

Note: Solving for $\displaystyle z$ gives $\displaystyle z = 2 \sin \frac{\theta}{2}$.

Do you mean:

$\displaystyle x = 1 + \cos \theta, y = \sin \theta, z = 2 \sin \frac{\theta}{2}, \theta = 0 \to 2 \pi$

If I try to solve for z using $\displaystyle x = 1 - \cos \theta, y = \sin \theta$, then I get to $\displaystyle z = 2\sqrt{\frac{1 + \cos \theta}{2}}$, and I cannot use a half-angle formula.

But if I solve for z using $\displaystyle x = 1 + \cos \theta, y = \sin \theta$, I get $\displaystyle z = 2 \sin \frac{\theta}{2}$ as you got.

That is a very helpful setup, but I have to wonder how you came up with that parametrization. I don't know how I could get such an answer without getting lucky enough to think to let $\displaystyle y = \sin \theta$ and then go from there.

Also, with that set up, it seems obvious enough that $\displaystyle 0 < \theta < 2 \pi$, but how can you tell which way to go? That's probably the toughest part of this type of problem for me.
• Apr 25th 2010, 12:10 PM
Jester
Quote:

Originally Posted by Redding1234
For the way I set it up, how could I figure out if $\displaystyle t$ is going from 0 to 2 or from 2 to 0? Is a picture necessary? I was having trouble drawing and labeling a picture of the curve.

Do you mean:

$\displaystyle x = 1 + \cos \theta, y = \sin \theta, z = 2 \sin \frac{\theta}{2}, \theta = 0 \to 2 \pi$

If I try to solve for z using $\displaystyle x = 1 - \cos \theta, y = \sin \theta$, then I get to $\displaystyle z = 2\sqrt{\frac{1 + \cos \theta}{2}}$, and I cannot use a half-angle formula.

But if I solve for z using $\displaystyle x = 1 + \cos \theta, y = \sin \theta$, I get $\displaystyle z = 2 \sin \frac{\theta}{2}$ as you got.

That is a very helpful setup, but I have to wonder how you came up with that parametrization. I don't know how I could get such an answer without getting lucky enough to think to let $\displaystyle y = \sin \theta$ and then go from there.

Also, with that set up, it seems obvious enough that $\displaystyle 0 < \theta < 2 \pi$, but how can you tell which way to go? That's probably the toughest part of this type of problem for me.

Yes, it's$\displaystyle x = 1 + \cos \theta$ (typo). So how did I come up with that. From the cylinder. If we re-write it as

$\displaystyle (x-1)^2 + y^2 = 1$ then let

$\displaystyle x - 1 = \cos \theta \; \text{and}\; y = \sin \theta$. As for direction. We know we're travelling around a circle. Pick two points for $\displaystyle \theta$ say $\displaystyle \theta = 0$ and $\displaystyle \theta = \pi/2$ to make sure we're travelling CCW around the circle.
• Apr 25th 2010, 04:11 PM
Redding1234
Ah, thanks!

I had one more problem that uses a similar concept, but rather than create a whole new thread, I thought I'd post it here.

I think I know how to do it, but I'd love it if someone could look at it and tell me how it looks.

Problem: Let $\displaystyle \omega = \frac{x}{x^2 + y^2}dx + \frac{y}{x^2 + y^2}dy$. If C is an arbitrary path from $\displaystyle (1,1)$ to $\displaystyle (2,2)$ not passing through the origin, calculate $\displaystyle \int_C{\omega}$.

I have the parameters of:

$\displaystyle x=1+t, y=1+t, t = 0 \to 1$

So $\displaystyle dx = dt, dy = dt$

Then $\displaystyle \int_{t=0}^1{\frac{1+t}{(1+t)^2 + (1+t)^2}dt + \frac{1+t}{(1+t)^2 + (1+t)^2}dt}$

$\displaystyle = \int_{t=0}^1{\frac{2(1+t)}{2(1+t)^2}dt} = \int_{t=0}^1{\frac{1}{(1+t)}dt} = ln(2)$

This seemed a little too easy to me for some reason. Is it okay to parametrize like this? If we're only considering where $\displaystyle t = 0 \to 1$, then that takes care of not passing through the origin, right?
• Apr 25th 2010, 04:50 PM
Jester
Quote:

Originally Posted by Redding1234
Ah, thanks!

I had one more problem that uses a similar concept, but rather than create a whole new thread, I thought I'd post it here.

I think I know how to do it, but I'd love it if someone could look at it and tell me how it looks.

Problem: Let $\displaystyle \omega = \frac{x}{x^2 + y^2}dx + \frac{y}{x^2 + y^2}dy$. If C is an arbitrary path from $\displaystyle (1,1)$ to $\displaystyle (2,2)$ not passing through the origin, calculate $\displaystyle \int_C{\omega}$.

I have the parameters of:

$\displaystyle x=1+t, y=1+t, t = 0 \to 1$

So $\displaystyle dx = dt, dy = dt$

Then $\displaystyle \int_{t=0}^1{\frac{1+t}{(1+t)^2 + (1+t)^2}dt + \frac{1+t}{(1+t)^2 + (1+t)^2}dt}$

$\displaystyle = \int_{t=0}^1{\frac{2(1+t)}{2(1+t)^2}dt} = \int_{t=0}^1{\frac{1}{(1+t)}dt} = ln(2)$

This seemed a little too easy to me for some reason. Is it okay to parametrize like this? If we're only considering where $\displaystyle t = 0 \to 1$, then that takes care of not passing through the origin, right?

New questions usually needs new threads. Something to think about in the future :).

Let me ask, the question says "any path" but you've chose a particular one. Why?
• Apr 25th 2010, 04:58 PM
Redding1234
I didn't know if it would be frowned upon to make two threads on the same day about a similar concept. Thanks for the advice though!

Is there a way to do the problem without using a particular path? My thinking was that it would be easiest to use that path. I'm not exactly sure how to calculate the integral without a specific path.