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Math Help - Area of a plane region

  1. #1
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    Area of a plane region

    This is an application of integrals:
    Heres the question
    Find the area of the region bounded by the curves:
    y^3 = x^2
    and
    x- 3y + 4 = 0

    The answer is 27/10 square units

    help and thanks ahead
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  2. #2
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    Krizalid's Avatar
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    Here's the solution on a PDF file.
    Attached Files Attached Files
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    This is an application of integrals:
    Heres the question
    Find the area of the region bounded by the curves:
    y^3 = x^2
    and
    x- 3y + 4 = 0

    The answer is 27/10 square units

    help and thanks ahead
    Hello,

    first you have to calculate the intercepts of both curves:

    y^3 = x^2 ===> y = (x^2)^(1/3)

    x-3y+4 = 0 ===> y = (1/3)x + 4/3

    Calculating the intercepts:

    (x^2)^(1/3) = (1/3)x + 4/3

    Raise both sides to the power of 3:

    x^2 = (1/27)x^3 + (4/9)x^2 + (16/9)x + (64/27)

    Multiply both sides by 27 and collect all terms at the LHS:

    x^3 - 15x^2 + 48x + 64 = 0. Now some miracle occurs (not really I only noticed that you can easily tranform the term so that (x-8) can be factored out)

    x^3 - 16x^2 + 64x + x^2 - 16x + 64 = 0

    x(x - 16x + 64) + (x - 16x + 64) = 0

    (x + 1)(x - 16x + 64) = 0 Thus you get the solutions: x = -1, x = 8

    Now you know the borders of the integral.

    The enclosed area is the integral of the difference of functions:

    d(x) = (x)^(1/3) - (1/3)x - (4/3)

    A = int[from -1 to 8]((x)^(1/3) - (1/3)x - (4/3))dx

    A = [from -1 to 8]((3/5)x^(5/3) - (1/6)x - (4/3)x)

    Plug in the values and you'll get:

    A= -(284/15) - (53/30) = -(207/10)

    Could it be that there is a typo in the answer key?

    EB

    I've attached a diagram.

    Edit: There must be a mistake in my calculations because this area isn't greater than 20 square units .... I can't find it ... sorry!
    Attached Thumbnails Attached Thumbnails Area of a plane region-eingeschl_flaeche.gif  
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  4. #4
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    Yep double checked it
    Thanks alot earboth & krizalid!
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