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Math Help - Area between three parabolas.

  1. #1
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    Area between three parabolas.

    I'm confused where to start with this. I have to calculate the area between three intersecting parabolas.

    y = x^2 - 8x + 16
    y = x^2 +8x +16
    y = 2 - 0.125x^2
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by eriiin View Post
    I'm confused where to start with this. I have to calculate the area between three intersecting parabolas.

    y = x^2 - 8x + 16
    y = x^2 +8x +16
    y = 2 - 0.125x^2

    to calculate area between them, first draw them, find intersecting points (to know limits of integration) ....
    then use double integral for calculating area and so on ...

    here are how do they look ( i zoom it a little so there doesn't show intersection points... that you should calculate )
    Attached Thumbnails Attached Thumbnails Area between three parabolas.-3functions.jpg  
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  3. #3
    Senior Member Educated's Avatar
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    Here's my attempt at it:

    1.  y = x^2 - 8x + 16

    2.  y = x^2 +8x +16

    3.  y = 2 - 0.125x^2

    First, find the interception point between 1. and 2. using simultanous equation, then find the interception point between 1. and 3.
    Integrate and solve the equation 1. using the interception points as your upper and lower. Call this area A.
    (I used the numbers closest to 0 as that is their central area of interception)
    This comes to:

    \displaystyle{A = \int_{-28/9}^0 (x^2 - 8x + 16) dx}


    Find the interception point between 2. and 3. and use your previous point of 1. and 2. interception point.
    Integrate and solve the equation using the interception points as your upper and lower. Call this area B.
    (I used the numbers closest to 0 as that is their central area of interception)
    This comes to:

    \displaystyle {B = \int_0^{28/9} (x^2 +8x +16 ) dx}


    Add A and B. This is the area under the parabolas 1. and 2.

    For parabola 3. use the same interception points, not 0.

    \displaystyle{ C = \int_{-28/9}^{28/9} (2 - 0.125x^2) dx}

    Minus the porabola C's area from A and B's total area:

    \mbox{Total area between the 3 porabola's} = A + B - C
    Last edited by Educated; August 20th 2010 at 11:36 PM.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Educated View Post
    Here's my go at it:

    1.  y = x^2 - 8x + 16

    2.  y = x^2 +8x +16

    3.  y = 2 - 0.125x^2

    First, find the interception point between 1. and 2. using simultanous equation, then find the interception point between 1. and 3. Use the lowest number co-ordinates
    Integrate and solve the equation 1. using the interception points as your upper and lower. Call this area A.
    (I used the numbers closest to 0 as that is their central area of interception)
    This comes to:

    \displaystyle{A = \int_{-28/9}^0 (x^2 - 8x + 16) dx}


    Find the interception point between 2. and 3. and use your previous point of 1. and 2. interception point.
    Integrate and solve the equation using the interception points as your upper and lower. Call this area B.
    (I used the numbers closest to 0 as that is their central area)
    This comes to:

    \displaystyle {B = \int_0^{28/9} (x^2 +8x +16 ) dx}


    Add A and B. This is the area under the porabolas 1. and 2.


    For porabola 3. use the same interception points, not 0.

    \displaystyle{ C = \int_{-28/9}^{28/9} (2 - 0.125x^2) dx}

    Minus the porabola 3. area from A and B's total area:

    \mbox{Total area between the 3 porabola's} = A + B - C
    there is no need for doing 3 integrals... it's way more easier to calculate (from as you wrote equations) intersection of 2. and 3. and do integration from that intersection to zero on x axis, and limits for y axis will be those to equations (from 3. to 1. equation)... and just do that same thing for right one... intersection of the 1. and 3. then do integration on x axis from zero to that intersection and integration on y axis will be from 3. to 1. equation...
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  5. #5
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    Not trying to sound rude or incredibly stupid, but yeKciM I have never used half of the stuff you've used in that equation. Like the Σ of cosh of tanh. Or if that's cos h and tan h, I don't know what h even stands for. I have fifteen huge exams coming up in the next two weeks so my minds just not working. D:
    I'll try doing the three integrals because that appears to make a little sense to me. Thanks for the help guys (:
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  6. #6
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by eriiin View Post
    Not trying to sound rude or incredibly stupid, but yeKciM I have never used half of the stuff you've used in that equation. Like the Σ of cosh of tanh. Or if that's cos h and tan h, I don't know what h even stands for. I have fifteen huge exams coming up in the next two weeks so my minds just not working. D:
    I'll try doing the three integrals because that appears to make a little sense to me. Thanks for the help guys (:
    hahahahahahaha
    sorry "eriiin" that have nothing to do with your question that sum and stuff is just cool way to write 1+1=2

    I'm really really sorry if that confused you in any way do it that way if it's more closer to you... my idea was just to do two double integrals... to get that region on the graph up there that i posted up there (post #2)

    if no one have give you any restrictions on how to do that (let's say that you must do it by double integrals) do it in any way that is more closer to you, and in way that you are sure that you know it
    good luck on exams
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  7. #7
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    Quote Originally Posted by eriiin View Post
    Not trying to sound rude or incredibly stupid, but yeKciM I have never used half of the stuff you've used in that equation. Like the Σ of cosh of tanh. Or if that's cos h and tan h, I don't know what h even stands for. I have fifteen huge exams coming up in the next two weeks so my minds just not working. D:
    I'll try doing the three integrals because that appears to make a little sense to me. Thanks for the help guys (:
    Most people have a "signature" quote or other information that is put at the end of each post. What you are talking about is yeKciM's "signature", not part of his answer.
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  8. #8
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    Oh god. Now I'm feeling likeeee incredibly stupid. Haha. I was like freaking out because I had no idea what you on about so thankyou so much for clearing that up ;
    I've never done double integrals but I've actually really never done any of this areas between two curves with integrals sooo I'll try my hardest to figure it out with the information you guys have provided. And again thankyouuuu!
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  9. #9
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    Sorry to bother again, however, in equation B where is uses x^2+8x+16 is it supposed to be the 1st equation with the +16 or is it supposed to be the -16 from equation 2?
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  10. #10
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    Nevermind what I asked. Sleep deprived and didn't read it properly.
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  11. #11
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    Sorry just curious, what program did you use to draw that graph? Using my calculator it draws it completely different.
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  12. #12
    Senior Member yeKciM's Avatar
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    hm... this one, in "Microsoft math"... but also occasional, (when have time to write code ) plot that in MATLAB 2010 or with Wolfram Mathematica....

    P.S. you should plot them by hand... just find few points (doesn't have to be 100% accurate lines) but as a sketch with exact points of intersection, direction ... that you can do very fast with pen and paper
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  13. #13
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    Quote Originally Posted by eriiin View Post
    Oh god. Now I'm feeling likeeee incredibly stupid.
    Just like the rest of us!!!

    Haha. I was like freaking out because I had no idea what you on about so thankyou so much for clearing that up ;
    I've never done double integrals but I've actually really never done any of this areas between two curves with integrals sooo I'll try my hardest to figure it out with the information you guys have provided. And again thankyouuuu!
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