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Math Help - Determine surface of area bounded by...

  1. #1
    Moo
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    Determine surface of area bounded by...

    Hello guys !

    I'd like to know the general method of doing when dealing with such texts :

    "find the area bounded by y=..... and y=......"

    or "find the volume of the area with a revolution around "...." axis"

    In fact, it'd be nice if you could give me examples of texts and their solving generalization, pleaaaaase
    I hope I'm not asking a too vague thing...

    Thanks ^^
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  2. #2
    Super Member wingless's Avatar
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    If you know how to evaluate double integrals it's easy.

    The area between f(x) and g(x) is

    \int_{b}^{a} \int_{g(x)}^{f(x)} ~dy~dx

    or

    \int_{d}^{c} \int_{g(y)}^{f(y)} ~dx~dy

    a and b are x coordinates, c and d are y coordinates of the intersection points of the curves.



    Example:
    Find the area of the circle x^2 + y^2 = 9.

    Solution (analytic, not geometric ;p)
    x^2 + y^2 = 9

    y^2 = 9-x^2

    This gives us two functions of x:
    f(x) = \sqrt{9-x^2}

    g(x) = -\sqrt{9-x^2}

    So we'll find the area between these curves.

    Find the x coordinates of the intersection points:

    \sqrt{9-x^2} = -\sqrt{9-x^2}

    x = 3,-3

    So the integral is,
    \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx

    \int_{-3}^{3} 2\sqrt{9-x^2}~dx

    x \sqrt{9-x^2}+9 \arcsin \left(\frac{x}{3}\right) ~|^{3}_{-3}

    9\pi
    Last edited by wingless; May 1st 2008 at 05:25 AM.
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  3. #3
    Moo
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    Hi !

    Quote Originally Posted by wingless View Post
    If you know how to evaluate double integrals it's easy.
    I'm not sure to get the whole meaning of "evaluate double integrals"... I know how to calculate them that's all :P

    The area between f(x) and g(x) is

    \int_{b}^{a} \int_{g(x)}^{f(x)} ~dx~dy

    or

    \int_{d}^{c} \int_{g(y)}^{f(y)} ~dy~dx

    a and b are y coordinates, c and d are x coordinates of the intersection points of the curves.
    Thanks for it, it's helping me a lot !



    And what about things introducing multiplication by \pi ? Is it always pi or can it be 2pi ? What is this about ?
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by Moo View Post
    And what about things introducing multiplication by \pi ? Is it always pi or can it be 2pi ? What is this about ?
    Is it about the volume of a revolution formula \pi \int^{a}_{b}f^2(x)~dx ?

    It's always \pi in this formula because it comes from the area of a circle \pi\cdot r^2.

    If you want to find the volume of a revolution of \alpha degrees, you can use it like,

    \frac{\alpha}{2\pi}\cdot\pi \int^{a}_{b}f^2(x)~dx
    (if alpha is in radian)

    \frac{\alpha}{360}\cdot\pi \int^{a}_{b}f^2(x)~dx
    (if alpha is in degrees)
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  5. #5
    Moo
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    Oh !
    And how are defined a and b ? Intersection of the curve with the axis ?
    I didn't look further, but is there an importance in the order of a and b ?
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by Moo View Post
    Oh !
    And how are defined a and b ? Intersection of the curve with the axis ?
    I didn't look further, but is there an importance in the order of a and b ?

    No, intersection of the curves f(x) and g(x). Set f(x) = g(x) and you'll get two x values, these are a and b. Normally, b must be the least one and a must be the bigger one. But it doesn't matter if you change their order, you'll get the negative of the same value.

    \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx = 9\pi

    \int_{3}^{-3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx = -9\pi

    Also, as a general rule,

    \int_{b}^{a}f(x)~dx = - \int_{a}^{b}f(x)~dx
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  7. #7
    Moo
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    I'm gonna be lectured because I didn't quote...

    Quote Originally Posted by wingless View Post
    Is it about the volume of a revolution formula \pi \int^{a}_{b}f^2(x)~dx ?
    I was asking about that : what are a and b ? Yep, it was about this formula
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  8. #8
    Super Member wingless's Avatar
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    Sorry, I don't have time to explain them at the moment, but these sites explain it as well:

    http://www.maths.abdn.ac.uk/~igc/tch/ma1002/int/node22.html
    Volume of A Solid of Revolution
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  9. #9
    Moo
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    Hey, thanks ! This is really enough for me and I think I understand where those weird formulas come from

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  10. #10
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    Quote Originally Posted by wingless View Post

    So the integral is,
    \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx
    It's actually easier by transforming this into a polar transform. So, \int_{-3}^{3}{\int_{-\sqrt{9-x^{2}}}^{\sqrt{9-x^{2}}}{dy}\,dx}=\int_{0}^{2\pi }\!{\int_{0}^{3}{r\,dr}\,d\theta }=9\pi .

    Or by considering \int_{-3}^{3}{\sqrt{9-x^{2}}\,dx}, which we're computing the area of a semicirle whose radius is 3.
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  11. #11
    Moo
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    It's not the problem of transforming into polar coordinates or finding tricks, I wanted the method and I have to admit, it was clear with what wingless did

    which we're computing the area of a semicirle whose radius is 3
    Solution (analytic, not geometric ;p)
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  12. #12
    Math Engineering Student
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    Sure. I read the last quote too. I posted just in case.
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  13. #13
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    Quote Originally Posted by Krizalid View Post
    Sure. I read the last quote too. I posted just in case.
    It's ok for that, thanks

    I really needed elements to understand problems that are sometimes posted here


    Do you know how to do when the axis around which there is the revolution is different from y=0 ?
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