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Math Help - integration/area problem

  1. #1
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    integration/area problem

    find the area of the loop defined by y^2=x^2(1-x)

    im quite confused with this question because im not really sure what the "loop" really refers to, i graphed the curve and assumed that the loop referred to when x is between 1 and 0. I then proceeded to find y in terms of x which i THINK becomes y=x[sqrt(1-x)]. After this, i integrated the equation using u substitution but my answer was 4/15 and not, as the answer says, 8/15.

    If anyone could direct me on where i went wrong, that would be a great help. thank you for your time and effort.
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  2. #2
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    Quote Originally Posted by iiharthero View Post
    find the area of the loop defined by y^2=x^2(1-x)

    im quite confused with this question because im not really sure what the "loop" really refers to, i graphed the curve and assumed that the loop referred to when x is between 1 and 0. I then proceeded to find y in terms of x which i THINK becomes y=x[sqrt(1-x)]. After this, i integrated the equation using u substitution but my answer was 4/15 and not, as the answer says, 8/15.

    If anyone could direct me on where i went wrong, that would be a great help. thank you for your time and effort.

    Everything you did is fine, but forgot that the given curve goes both above AND below the x-axis and, in fact, it is symetric wrt it, so you only calculated half the actual area: the one over the x-axis!

    Tonio
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  3. #3
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    thank you for your help but im a little unsure on what this particular graph looks like, because i dont have any symmetry on mine T.T could you give me a few pointers on how to graph it?
    thanks again
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  4. #4
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    Quote Originally Posted by iiharthero View Post
    thank you for your help but im a little unsure on what this particular graph looks like, because i dont have any symmetry on mine T.T could you give me a few pointers on how to graph it?
    thanks again

    Simply take the square root of the equation to put it in function-like form:

    y^2=x^2(1-x) \Longrightarrow y=\pm x \sqrt{1-x}

    and now just graph the plus sign, which ressembles the positive part of the parabola -x(x-1).

    Finally, repeat in a mirror-like fashion the graph under the x-axis and you're done.

    Tonio
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