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Math Help - Area between two curves problem

  1. #1
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    Area between two curves problem

    Provided:

    y = 5 - x

    x = (1/6)(y-5)^2

    How do I know to solve in terms of y or x?

    So far I've tried it in terms of y and found the points of intersection to be -1, 5.. I tried evaluating the integral of \int(5-y)-(1/6(y-5)^2), with the limits -1 to 5, but was marked incorrect... Any suggestions? Should it be in terms of x ?

    Thanks,

    John
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  2. #2
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    You're integrating with respect to y, you need to find the limits along the y axis. You're using the limits along the x-axis. Try integrating from 0 to 6.
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  3. #3
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    Quote Originally Posted by JohnM25 View Post
    Provided:

    y = 5 - x

    x = (1/6)(y-5)^2

    How do I know to solve in terms of y or x?

    So far I've tried it in terms of y and found the points of intersection to be -1, 5.. I tried evaluating the integral of \int(5-y)-(1/6(y-5)^2), with the limits -1 to 5, but was marked incorrect...
    That integral \displaystyle\int_{-1}^5\!\!\bigl(5-y-\tfrac16(y-5)^2\bigr)dy looks correct to me. Maybe you evaluated it wrongly. Did you get the answer 6?
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    Quote Originally Posted by Opalg View Post
    That integral \displaystyle\int_{-1}^5\!\!\bigl(5-y-\tfrac16(y-5)^2\bigr)dy looks correct to me. Maybe you evaluated it wrongly. Did you get the answer 6?
    You're right... I think I made a mistake in the integration of 1/6(y-5)^2.. Thank you
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  5. #5
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    Quote Originally Posted by Opalg View Post
    That integral \displaystyle\int_{-1}^5\!\!\bigl(5-y-\tfrac16(y-5)^2\bigr)dy looks correct to me. Maybe you evaluated it wrongly. Did you get the answer 6?
    Isn't he integrating from y=0 to y=6 ? I graphed the functions and they intersect at (-1,6) and (5,0). You can't plug in x values for y.
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  6. #6
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    Quote Originally Posted by adkinsjr View Post
    Isn't he integrating from y=0 to y=6 ? I graphed the functions and they intersect at (-1,6) and (5,0). You can't plug in x values for y.
    You seem to have the x and y coordinates the wrong way round. The points of intersection are (0,5) and (6,–1).
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  7. #7
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    Quote Originally Posted by Opalg View Post
    You seem to have the x and y coordinates the wrong way round. The points of intersection are (0,5) and (6,1).
    LOL, yeah ok, I graphed the functions on my calculator, and it was giving (5,0) and (-1,6) as the intersection points, but I forgot that I interchanged x and y when plugging it in:

    y_1=5-x
    y_2=1/6(x-5)^2

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