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Math Help - Line integrals - how to determine limits

  1. #1
    Junior Member Pinsky's Avatar
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    Line integrals - how to determine limits

    Hy, i'm having problems with this example.

    \int_cxyds

    where c is one quarter of the ellipse

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

    The line integral itself isn't the problem, the problem tis how to define the limits of integration.

    I have the same problem with this example

    \int_c(2a-y)dx+xdy

    where c is an arc of the first bow of a trochoid
    x=a(t-\sin(t))
    y=a(1-\cos(t))
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  2. #2
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    Quote Originally Posted by Pinsky View Post
    Hy, i'm having problems with this example.

    \int_cxyds

    where c is one quarter of the ellipse

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

    The line integral itself isn't the problem, the problem tis how to define the limits of integration.

    I have the same problem with this example

    \int_c(2a-y)dx+xdy

    where c is an arc of the first bow of a trochoid
    x=a(t-\sin(t))
    y=a(1-\cos(t))
    In the first, I'd switch to a parametric form: x = a cos(t), y = b sin(t). Then the simplest quarter of the ellipse would give limits t = 0 to t = pi/2.

    In the second, have you tried sketching an arc of the first bow of the trochoid? It starts at t = 0. For what value of t does it end ......? By the way, did you notice that if a = 1 you have a cycloid? (This might help you answer my question.)
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  3. #3
    Junior Member Pinsky's Avatar
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    Let's say a have a line in 2D and i want to integrate over it in the first quadrant.
    If the function of a line it given explicitly y=f(x) than i can easily see for which values of x is the line in the first quadrant. But how do i do that if the line is given in parametric equations x=f(t) y=f(t)? Is there a method for doing this?

    Ask if you didn't understand my question.
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  4. #4
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    Quote Originally Posted by Pinsky View Post
    Let's say a have a line in 2D and i want to integrate over it in the first quadrant.
    If the function of a line it given explicitly y=f(x) than i can easily see for which values of x is the line in the first quadrant. But how do i do that if the line is given in parametric equations x=f(t) y=f(t)? Is there a method for doing this?

    Ask if you didn't understand my question.
    Yes there is.

    BUT it will be easier to show you by using an example. So ....... what line of the form y = f(x) do you have in mind?
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  5. #5
    Junior Member Pinsky's Avatar
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    Let's say we use a cykloide, and it is already given as a parametric equation x=f(t) y=f(t)

    x=a(t-\sin{t})
    y=a(1-\cos{t})

    So now i have to calculate a line integral
    \int_c(2a-y)dx+xdy
    Where c is the first arc of the cycloide (the first example i gave, it turned to be a cycloide after all).

    This is how i do it:
    First
    dx=(a-a\cos{t})dt
    dy=a\sin{t}dt

    So now i include all of that into the integral. But i don't know how to determine the borders. It can be seen from the picture that if i want to integrate until the first arc, i should use limits from 0 to 2*r*pi or in my case to 2*a*pi. But those would be the limits if i would integrate with dx.
    So how do i find the limits when integrating with dt?
    Attached Thumbnails Attached Thumbnails Line integrals - how to determine limits-cikloida.gif  
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  6. #6
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    Quote Originally Posted by Pinsky View Post
    Let's say we use a cykloide, and it is already given as a parametric equation x=f(t) y=f(t)

    x=a(t-\sin{t})
    y=a(1-\cos{t})

    So now i have to calculate a line integral
    \int_c(2a-y)dx+xdy
    Where c is the first arc of the cycloide (the first example i gave, it turned to be a cycloide after all).

    This is how i do it:
    First
    dx=(a-a\cos{t})dt
    dy=a\sin{t}dt

    So now i include all of that into the integral. But i don't know how to determine the borders. It can be seen from the picture that if i want to integrate until the first arc, i should use limits from 0 to 2*r*pi or in my case to 2*a*pi. But those would be the limits if i would integrate with dx.
    So how do i find the limits when integrating with dt?
    \int_c (2a-y) \, dx + x \, dy = \int_{t = t_1}^{t = t_2} \left( (2a-y) \, \frac{dx}{dt} + x \, \frac{dy}{dt} \right) \, dt = .......

    where I leave it to you to sub in x=a(t-\sin{t}), y=a(1-\cos{t}), \frac{dx}{dt} = a(1 - \cos{t}), \frac{dy}{dt} = a\sin{t}, t_1 = 0 and t_2 = 2\pi (NOT 2 a \pi) and then evaluate the resulting definite integral.
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  7. #7
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    In parametric, one arch of the cycloid.

    x = a(t-sint), \;\ y=a(1-cost)

    (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=[a(1-cost)]^{2}+[asint]^{2}=2a^{2}(1-cost)=4a^{2}sin^{2}(\frac{t}{2})

    L=\int_{0}^{2\pi}2asin(\frac{t}{2})dt

    I included a diagram. Hope it helps. Perhaps you already know it, though.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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    Quote Originally Posted by galactus View Post
    In parametric, one arch of the cycloid.

    x = a(t-sint), \;\ y=a(1-cost)

    (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=[a(1-cost)]^{2}+[asint]^{2}=2a^{2}(1-cost)=4a^{2}sin^{2}(\frac{t}{2})

    L=\int_{0}^{2\pi}2asin(\frac{t}{2})dt

    I included a diagram. Hope it helps. Perhaps you already know it, though.
    But just to make it clear, you've set up the integral for calculating the length of the arch (an interesting and useful result btw ), not the line integral that Pinsky posted. Right?
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    DUH, Thanks MrF. I'll leave it anyway. May help.

    BTW, isn't the cycloid also known as the 'brachistochrone'?. Wasn't this the problem one of the Bernoulli's posed to Newton and he solved it rather quickly, according to the story?. The problem was to determine the best shape for an object, such as a bearing might roll from a point P to a point Q, not directly below it, the fastest. This is the shape Newton came up with.
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  10. #10
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    Quote Originally Posted by galactus View Post
    DUH, Thanks MrF. I'll leave it anyway. May help.

    BTW, isn't the cycloid also known as the 'brachistochrone'?. Wasn't this the problem one of the Bernoulli's [Mr F adds: Johann, in fact. You need a score card to keep a track of that family ] posed to Newton and he solved it rather quickly, according to the story?. The problem was to determine the best shape for an object, such as a bearing might roll from a point P to a point Q, not directly below it, the fastest. This is the shape Newton came up with.
    Indeed. But I think Bernoulli's own solution is the one that's mostly mentioned. The problem has a very interesting history, as does the cycloid (which had been known for at least 100 years prior to the problem Bernoulli posed).
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  11. #11
    Junior Member Pinsky's Avatar
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    But why did you take the limits of integration from 0 to 2\pi.
    How do you figure that out from the parametric equation, which is is my case a cycloide?

    Like in this example.
    \int_cxyds

    where c is one quarter of the ellipse

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
    Mr.fantastic said to take the limits of integration from 0 to \frac{\pi}{2}

    Why?
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  12. #12
    Junior Member Pinsky's Avatar
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    Found an almost analog problem, that deals whit double integrals and changeint the variables to polar cordinates.

    The task is to calculate an integral

    \int_D\int\sqrt{1-x^2-y^2}dxdy

    Where D is a half circle bounded by a circle (x-\frac{1}{2})^2+y^2=\frac{1}{4}
    and the x axis.

    This is a solved example from a book.
    So the first thing they do is transform the formula of the circuit into polar coordinates.
    So

    (x-\frac{1}{2})^2+y^2=\frac{1}{4}\Rightarrow r=\cos{\varphi}

    So in the end they get:


    \int_{0}^{\frac{\pi}{2}}\int_{0}^{\cos{\varphi}}\s  qrt{1-r^2}drd\varphi

    I've drawn the half circle in the first quadrant. An then again my question, how where the limits 0 to \frac{\pi}{2} defined?
    Why aren't the limits from 0 to \pi? That's a half circle then.
    Attached Thumbnails Attached Thumbnails Line integrals - how to determine limits-kruznica.png  
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    Quote Originally Posted by Pinsky View Post
    But why did you take the limits of integration from 0 to 2\pi.
    How do you figure that out from the parametric equation, which is is my case a cycloide?

    Like in this example.


    Mr.fantastic said to take the limits of integration from 0 to \frac{\pi}{2}

    Why?
    The easiest way for you to see this is to actually substitute values of t into the parametric equations defining each curve and then plot the points .....

    Eg. What do you get if you substitute values of t from 0 to pi/2 into the parametric equations defining the ellipse ..... What happens as you go past t = pi/2 .....

    (If you have access to a graphics calculator you can draw curves in parametric mode and use the trace feature)

    It looks to me like you don't have a lot of experience with curves defined in parametric form. It will be well worth your academic while to jump onto that particular learning curve.
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  14. #14
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    Quote Originally Posted by Pinsky View Post
    Found an almost analog problem, that deals whit double integrals and changeint the variables to polar cordinates.

    The task is to calculate an integral

    \int_D\int\sqrt{1-x^2-y^2}dxdy

    Where D is a half circle bounded by a circle (x-\frac{1}{2})^2+y^2=\frac{1}{4}
    and the x axis.

    This is a solved example from a book.
    So the first thing they do is transform the formula of the circuit into polar coordinates.
    So

    (x-\frac{1}{2})^2+y^2=\frac{1}{4}\Rightarrow r=\cos{\varphi}

    So in the end they get:


    \int_{0}^{\frac{\pi}{2}}\int_{0}^{\cos{\varphi}}\s  qrt{1-r^2}drd\varphi

    I've drawn the half circle in the first quadrant. An then again my question, how where the limits 0 to \frac{\pi}{2} defined?
    Why aren't the limits from 0 to \pi? That's a half circle then.
    On the circle (x-\frac{1}{2})^2+y^2=\frac{1}{4} defining the region of integration, the point (1, 0) corresponds to the polar coords [1, 0] (capisce?). The point (0.5, 0.5) corresponds to \left[\frac{1}{\sqrt{2}}, \frac{\pi}{4}\right] (capisce?). The point (0, 0) corresponds to [0, pi/2] (capisce?).

    I hope you can see that as you move around the circle from (1, 0) to (0.5, 0.5), r goes from 1 to \frac{1}{\sqrt{2}} and \varphi goes from 0 to \frac{\pi}{4}.

    And I hope you can see that as you move around the circle from (0.5, 0.5) to (0, 0), r goes from \frac{1}{\sqrt{2}} to 0 and \varphi goes from \frac{\pi}{4} to \frac{\pi}{2}.

    Indeed, if you express the circle (x-\frac{1}{2})^2+y^2=\frac{1}{4}\Rightarrow x^2 - x + y^2 = 0 in polar coordinates, you get r = \cos \varphi ...... If you draw a graph of r versus \varphi you will clearly see r changing from 1 to 0 as \varphi goes from 0 to \frac{\pi}{2}. And obviously r = 1 corresponds to the point (1, 0) and r = 0 corresponds to the origin (0, 0).

    What happens as \varphi goes from \frac{\pi}{2} to \pi .....? What happens as \varphi increases from \pi ....?

    Again, you need to jump onto the bus going to parametric city. Make sure it's an express.
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