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Math Help - Area between two curves

  1. #1
    Member Marconis's Avatar
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    Area between two curves

    We just finished up area between two curves and volume of a solid. I've been having some difficulties with it. This was a homework problem two nights ago, and I thought I understood it (even after doing it in class), but I don't.

    The question tells you to find the area between:

    4x+y^2=12 and x=y

    So I solve for x.

    4x=-y^2+12

    x=\frac{-y^2+12}{4}

    Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?

    The answer was eventually shown as (before computation):

    A= \int^{2} [(y) - (\frac{-y^2+12}{4})]dy

    I forgot how to put the lower part of the integral in there, sorry.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Marconis View Post
    We just finished up area between two curves and volume of a solid. I've been having some difficulties with it. This was a homework problem two nights ago, and I thought I understood it (even after doing it in class), but I don't.

    The question tells you to find the area between:

    4x+y^2=12 and x=y

    So I solve for x.

    4x=-y^2+12

    x=\frac{-y^2+12}{4}

    Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?

    The answer was eventually shown as (before computation):

    A= \int^{2} [(y) - (\frac{-y^2+12}{4})]dy

    I forgot how to put the lower part of the integral in there, sorry.
    4x + y^2 = 12

    and
    y = x

    So put y in for the x value:
    4y + y^2 = 12

    Solving for y gives you -6 and 2 as I'm sure you know. Now, y = x, so the x values at the points of intersection are the same as the y values. So the points of intersection are (-6., -6) and (2, 2).

    -Dan
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  3. #3
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    Quote Originally Posted by Marconis View Post

    Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?
    Take \displaystyle y = \frac{12-y^2}{4}

    ...

    \displaystyle (y-2)(y+6)=0
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  4. #4
    Member Marconis's Avatar
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    Ah, okay. I was unsure if you could do that. Lol. *Sigh* math.

    Thanks!
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  5. #5
    Member Marconis's Avatar
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    Sorry, one more thing...

    I can't seem to recognize this graphically...any assistance on this?
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  6. #6
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    Sketch y=x and y=\sqrt{12-4x} on the same axis, you are looking for the area in between.
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  7. #7
    Member Marconis's Avatar
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    Got it, thank you.
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  8. #8
    Member Marconis's Avatar
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    NEVERMIND, figured it out.
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