Results 1 to 7 of 7

Math Help - Need some help with a line integral problem...

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    30

    Need some help with a line integral problem...

    Problem: Let C be the curve of intersection of the upper hemisphere x^2 + y^2 + z^2 = 4, z\geq0 and the cylinder x^2 + y^2 = 2x, oriented counterclockwise as viewed from high above the xy-plane. Evaluate \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:

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

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

    z = t

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

    \int_{t=a}^b {[\frac{t\sqrt{4 - t^2}}{2} * -t}  + 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; April 25th 2010 at 10:51 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by Redding1234 View Post
    Problem: Let C be the curve of intersection of the upper hemisphere x^2 + y^2 + z^2 = 4, z\geq0 and the cylinder x^2 + y^2 = 2x, oriented counterclockwise as viewed from high above the xy-plane. Evaluate \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:

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

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

    z = t

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

    \int_{t=a}^b {[\frac{t\sqrt{4 - t^2}}{2} * -t} + 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 y there's a \pm to consider. Since your primary variable is z = t, then  t is between 0 and 2 but be careful on which way you go (remember, you gotta transverse CCW around the curve).

    I think the parameterization

     <br />
x = 1 - \cos \theta, y = \sin \theta, \theta = 0 \to 2 \pi<br />

    is better and much more natural.

    Note: Solving for z gives z = 2 \sin \frac{\theta}{2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    For the way I set it up, how could I figure out if 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

     <br />
x = 1 - \cos \theta, y = \sin \theta, \theta = 0 \to 2 \pi<br />

    is better and much more natural.

    Note: Solving for z gives z = 2 \sin \frac{\theta}{2}.
    Do you mean:

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

    But if I solve for z using x = 1 + \cos \theta, y = \sin  \theta, I get 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 y = \sin \theta and then go from there.

    Also, with that set up, it seems obvious enough that 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by Redding1234 View Post
    For the way I set it up, how could I figure out if 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:

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

    But if I solve for z using x = 1 + \cos \theta, y = \sin \theta, I get 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 y = \sin \theta and then go from there.

    Also, with that set up, it seems obvious enough that 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  x = 1 + \cos \theta (typo). So how did I come up with that. From the cylinder. If we re-write it as

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

    x - 1 = \cos \theta \; \text{and}\; y = \sin \theta. As for direction. We know we're travelling around a circle. Pick two points for \theta say \theta = 0 and \theta = \pi/2 to make sure we're travelling CCW around the circle.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    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 \omega = \frac{x}{x^2 + y^2}dx + \frac{y}{x^2 + y^2}dy. If C is an arbitrary path from (1,1) to (2,2) not passing through the origin, calculate \int_C{\omega}.

    I have the parameters of:

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

    So dx = dt, dy = dt

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

     = \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 t = 0 \to 1, then that takes care of not passing through the origin, right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    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 \omega = \frac{x}{x^2 + y^2}dx + \frac{y}{x^2 + y^2}dy. If C is an arbitrary path from (1,1) to (2,2) not passing through the origin, calculate \int_C{\omega}.

    I have the parameters of:

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

    So dx = dt, dy = dt

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

     = \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 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?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Line integral problem.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 31st 2010, 11:47 AM
  2. problem with line integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2010, 03:32 AM
  3. Line integral problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 4th 2009, 07:24 AM
  4. Line Integral Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2009, 08:02 AM
  5. Line Integral Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 21st 2008, 12:35 AM

Search Tags


/mathhelpforum @mathhelpforum