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Math Help - define integral of the circle y=sqrt(49-x^2)1/2

  1. #1
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    Question define integral of the circle y=sqrt(49-x^2)

    i have this equation and i wanna the upper limit of it cuz i know the area
    under the curve here it is
    Last edited by nawartamawi; April 17th 2010 at 10:41 AM.
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  2. #2
    Super Member Deadstar's Avatar
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    Set x = 7\sin(u).

    Then dx = 7\cos(u)du.

    So our integral becomes...

    \int_0^{\arcsin(\tfrac{1}{7}p)} \sqrt{49 - 49\sin^2(u)} 7\cos(u) du

    = \int_0^{\arcsin(\tfrac{1}{7}p)} 49\sqrt{\cos^2(u)} \cos(u) du

    = \int_0^{\arcsin(\tfrac{1}{7}p)} 49\cos^2(u) du

    = \bigg{[} 49 \bigg{(} \frac{u}{2} + \frac{1}{4}\sin(2u) \bigg{)}\bigg{]}_0^{\arcsin(\tfrac{1}{7}p)}

    = 49 \bigg{(} \frac{\arcsin(\tfrac{1}{7}p)}{2} + \frac{1}{4}\sin(2\arcsin(\tfrac{1}{7}p)) \bigg{)}

    Posting working as I go along...
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by nawartamawi View Post
    i have this equation and i wanna the upper limit of it cuz i know the area
    under the curve here it is
    Evaluate the integral and solve for P

    let x = 7sin \theta

    So our bounds become

    upper:  \arcsin \frac{p}{7}

    lower:  0

    dx = 7cos \theta d \theta

    We will leave the limits for now.

    \int \sqrt{7^2 - 7^2 sin^2 \theta} (7cos \theta) d \theta

    49 \int \sqrt{1 - 1 sin^2 \theta} (cos \theta) d \theta

    49 \int \sqrt{cos^2 \theta } (cos \theta) d \theta

    49 \int cos^2 d \theta

    49 \int \frac{1-cos2x}{2} d \theta

    \frac{49}{2} \int d \theta - \frac{49}{2} \int cos2x d \theta

    \frac{49}{2} \theta + \frac{49}{4} sin2 \theta

    Remember

    \theta=\arcsin \frac{x}{7}

    \frac{49}{2} \arcsin \frac{x}{7}+ \frac{49}{4} sin(2 \arcsin \frac{x}{7})

    This is evaluated from 0--P so we have

    \frac{49}{2} \arcsin \frac{p}{7}+ \frac{49}{4} sin(2 \arcsin \frac{p}{7})

    Let us note that any combination of n \pi will make sin = 0 but we will take n=0 for simplicity.

    \frac{49}{2} \arcsin \frac{p}{7}+ \frac{49}{4} sin(2 \arcsin \frac{p}{7}) = \frac{49 \pi}{12}

    Solve for P
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  4. #4
    Super Member Deadstar's Avatar
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    Ugg ok scrap that use this instead...

    \int \cos^2(x) dx = \frac{1}{2}\cos(x)\sin(x)+\frac{1}{2}x

    So we get instead...

    = \int_0^{\arcsin(\tfrac{1}{7}p)} 49\cos^2(u) du = \bigg{[} \frac{49}{2}\cos(u)\sin(u)+\frac{49}{2}u \bigg{]}_0^{\arcsin(\tfrac{1}{7}p)}

    = \frac{49}{2}\cos(\arcsin(\tfrac{1}{7}p))\sin(\arcs  in(\tfrac{1}{7}p))+\frac{49}{2}\arcsin(\tfrac{1}{7  }p)

    =  \frac{49}{2} \cdot \frac{1}{7} \cdot \sqrt{49-p^2} \cdot \frac{p}{7}+\frac{49}{2}\arcsin(\tfrac{1}{7}p)

    = \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p)
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  5. #5
    Super Member Deadstar's Avatar
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    Then...

    Set this equal to...

    \frac{49\pi}{12}...

    To get this hellish thing to solve.

    = \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) =  \frac{49\pi}{12}

    => = 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi

    ...

    Using Maple now since I don't know how to simplify this off the top of my head you get
    <br />
p=1.854524594
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  6. #6
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Deadstar View Post
    Then...

    Set this equal to...

    \frac{49\pi}{12}...

    To get this hellish thing to solve.

    = \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) = \frac{49\pi}{12}

    => = 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi

    ...

    Using Maple now since I don't know how to simplify this off the top of my head you get
    <br />
p=1.854524594
    We must have both missed something because there is no way that can be done by hand...
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  7. #7
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    thank you very much i really apreciate your work
    and there is somthing else
    iwould like to ask you if we plot a graph of the last equation we have
    = 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi
    shall we get somthing
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  8. #8
    Super Member Deadstar's Avatar
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    Quote Originally Posted by nawartamawi View Post
    thank you very much i really apreciate your work
    and there is somthing else
    iwould like to ask you if we plot a graph of the last equation we have
    = 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi
    shall we get somthing
    See attached images.

    First one is the graph of

    6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) - 49\pi

    We want to know where this one crosses the x-axis.

    Second one the graphs of of...

    6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) \textrm{ and } 49\pi

    With the blue line representing 49\pi.

    We want to see where they cross.

    This value is the p value I posted above. Maybe Opalg or Krizalid could give you a better answer but this is the best I can do for now.
    Attached Thumbnails Attached Thumbnails define integral of the circle y=sqrt(49-x^2)1/2-asdffa.jpg   define integral of the circle y=sqrt(49-x^2)1/2-asdfg.jpg  
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  9. #9
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    Quote Originally Posted by Deadstar View Post
    Then...

    Set this equal to...

    \frac{49\pi}{12}...

    To get this hellish thing to solve.

    = \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) =  \frac{49\pi}{12}

    => = 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi

    ...




    Using Maple now since I don't know how to simplify this off the top of my head you get
    <br />
p=1.854524594
    Quote Originally Posted by Deadstar View Post
    See attached images.

    First one is the graph of

    6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) - 49\pi

    We want to know where this one crosses the x-axis.

    Second one the graphs of of...

    6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) \textrm{ and } 49\pi

    With the blue line representing 49\pi.

    We want to see where they cross.

    This value is the p value I posted above. Maybe Opalg or Krizalid could give you a better answer but this is the best I can do for now.



    its cool thank very much now its so clear

    but if someday you could solve it by hand post it to me
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  10. #10
    Super Member Deadstar's Avatar
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    Quote Originally Posted by AllanCuz View Post
    We must have both missed something because there is no way that can be done by hand...
    Maybe... But perhaps this was just something the OP wanted to know rather than a set tutorial/homework question with a 'nice' answer.
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  11. #11
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    Quote Originally Posted by nawartamawi View Post
    i have this equation and i wanna the upper limit of it cuz i know the area
    under the curve here it is
    As the curve is a semicircle, the integral is the sum of sector area and a triangle area.

    sin\left(\frac{{\pi}}{2}-\theta\right)=\frac{P}{7}

    \frac{{\pi}}{2}-\theta=sin^{-1}\left(\frac{P}{7}\right)

    The integral is

    \frac{49}{2}sin^{-1}\left(\frac{P}{7}\right)+\frac{P}{2}\sqrt{49-P^2}

    then use the intersection of curves and lines or other methods for P.
    Attached Thumbnails Attached Thumbnails define integral of the circle y=sqrt(49-x^2)1/2-semicircle.jpg  
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