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Math Help - Can someone check this - Area between curves

  1. #1
    Member mybrohshi5's Avatar
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    Can someone check this - Area between curves

    Find the area between  y = 20-x^2 and y=x^2-12

    I found the integral to be

    \int_{-4}^{4}[(20-x^2)-(x^2-12)]dx

    and after evaluating that i found the area to be 170.66667

    Does this seem correct?

    Thank you?
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  2. #2
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    better if along y-axis

    Quote Originally Posted by mybrohshi5 View Post
    Find the area between  y = 20-x^2 and y=x^2-12

    I found the integral to be

    \int_{-4}^{4}[(20-x^2)-(x^2-12)]dx

    and after evaluating that i found the area to be 170.66667

    Does this seem correct?

    Thank you?
    y=x^2-12 is a parabola pointing upwards with vertex at (0,-12) while the other is a parabola pointing downwards with vertex at (0,20). both of them meet at(+-4,4).

    Now integrating along the x-axis will lead to complications. better will be to integrate along y-axis.then the integral will be from 20 to -12and the expression will be as in the attachment.
    Attached Thumbnails Attached Thumbnails Can someone check this - Area between curves-integral.bmp  
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  3. #3
    Member mybrohshi5's Avatar
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    I just tried what you attached and i got 0 for my answer.

    Does that seem right?

    Thanks for the help
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  4. #4
    Member mybrohshi5's Avatar
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    Looking at my original graph again of y=20-x^2 and y=x^2-12 i notice that there is more area above the x axis than below so it doesnt seem right to me that the area is 0. Just from looking at the graph it looks like the area should maybe be a positive number????
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  5. #5
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    unable to find the mistake

    Quote Originally Posted by mybrohshi5 View Post
    Looking at my original graph again of y=20-x^2 and y=x^2-12 i notice that there is more area above the x axis than below so it doesnt seem right to me that the area is 0. Just from looking at the graph it looks like the area should maybe be a positive number????
    you are right. the answer comes out to be 0.which is wrong. i am not able to find out my mistake. either there is a mistake in putting the limits(which is less probable) or i put the value of x instead of y to integrate. sorry for the mistake.
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  6. #6
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    got the mistake!

    Quote Originally Posted by mybrohshi5 View Post
    Looking at my original graph again of y=20-x^2 and y=x^2-12 i notice that there is more area above the x axis than below so it doesnt seem right to me that the area is 0. Just from looking at the graph it looks like the area should maybe be a positive number????
    mistake in putting the limits:

    there will be three integrals:

    I1=integration(20,4)sqrt(y+12)dy
    I2=integration(0,4)sqrt(20-y)dy
    I3=integration(0,12)sqrt(20-y)dy

    area=I1+I2+I3.
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  7. #7
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    I don't see anything wrong with just doing
    \int_{-4}^{4}[(20-x^2)-(x^2-12)]dx
    as you say in your first post.

    \int_{-4}^4 (32- 2x^2) dx= \left[32x- 4/3 x^3\right]_{-4}^4
    which equals 85 and 2/3, not 170 and 2/3.
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  8. #8
    Member mybrohshi5's Avatar
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    I believe you made a mistake on your integral. I think it should be


    <br />
\int_{-4}^4 (32- 2x^2) dx= \left[32x- (2/3) x^3\right]_{-4}^4<br />

    which equals 170 and 2/3

    Thank you for the help
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  9. #9
    MHF Contributor

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    Yes, of course!
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