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Thread: Need some help with a line integral problem...

  1. #1
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    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.
    Last edited by Redding1234; Apr 25th 2010 at 10:51 PM.
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    Quote Originally Posted by Redding1234 View Post
    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 $\displaystyle z = 2 \sin \frac{\theta}{2}$.
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    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 View Post
    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.
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    Quote Originally Posted by Redding1234 View Post
    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.
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    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?
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    Quote Originally Posted by Redding1234 View Post
    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?
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  7. #7
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    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.
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