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Math Help - Problem 15: Area between two curves

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    Problem 15: Area between two curves

    Let R be the region in the upper half plane between the curves 2x^4 + y^4 + y = 2 and x^4 + 8y^4 + y = 1.

    Find: \iint _R \frac{1}{(x^2 + 1)y}~dy~dx

    -Dan
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    Forum Admin topsquark's Avatar
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    Re: Problem 15: Area between two curves

    Hint: Is there a relationship between the two boundary curves?

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    Re: Problem 15: Area between two curves

    Let's start by looking at the hint and describe the curves.
    1) 2x^4 + y^4 + y = 2

    2) x^4 + 8y^4 + y = 1

    Notice that equation 2) can be rewritten as 2x^4 + (2y)^4 + (2y) = 2 so equation 2) is just equation 1) with a vertical stretch factor of 2.

    I am going to represent equation 1) as y = f(x) and equation 2) as y = 2 f(x).

    A couple of important facts about curve 1). Note that f(-x) = f(x), so f(x) is an even function. Also note that f(x) has two real zeros: x = \pm 1. And finally, if we look at y = f(x) in quadrant 1 we find that f(x) is monotonically decreasing on [0, 1], so f(x) is one to one in Quadrant 1. (And thus is a function on [-1, 1].) Similar comments obviously apply to y = 2 f(x), and it is trivial to show that the only intersection points of equations 1) and 2) are at x = \pm 1.

    Our region of integration can thus be characterized as (2 f(x) - f(x))dx integrated over the interval [-1, 1].

    Thus
    \iint _{R} \frac{1}{y(x^2 + 1)}~dy~dx = \int _{-1}^{1} \int _{f(x)} ^{2f(x)} \frac{1}{(x^2 + 1)y}~dy~dx

    The integration is now simple. I'll sketch it briefly:
    \iint _{R} \frac{1}{(x^2 + 1)y}~dy~dx = \int _{-1}^{1} \int _{f(x)} ^{2f(x)} \frac{1}{(x^2 + 1)y}~dy~dx

    = \int_{-1}^{1} \frac{1}{x^2 + 1} \left [ \int_{f(x)}^{2f(x)} \frac{1}{y}~dy \right ] dx

    = \int_{-1}^{1} \frac{1}{x^2 + 1} \left ( ln(2 f(x)) - ln(f(x) \right ) ~dx

    = ln(2) \int_{-1}^{1} \frac{1}{x^2 + 1}~dx

    = \ln(2) \left ( tan^{-1}(1) - tan^{-1}(-1) \right )

    = \frac{\pi}{2} ln(2)

    -Dan
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    Re: Problem 15: Area between two curves

    There are four points of intersection:

    (-1,0), (1,0), \left( -\sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right) \approx (-1.04428,-0.405480), \left( \sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right)\approx (1.04428,-0.405480)

    Does your solution account for all four points of intersection?

    Edit: Never mind. I missed the part that said that R is restricted to the upper half-plane.
    Last edited by SlipEternal; July 29th 2014 at 11:59 AM.
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    Re: Problem 15: Area between two curves

    Quote Originally Posted by SlipEternal View Post
    There are four points of intersection:

    (-1,0), (1,0), \left( -\sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right) \approx (-1.04428,-0.405480), \left( \sqrt[4]{1+7\cdot 15^{-4/3} }, -15^{-1/3} \right)\approx (1.04428,-0.405480)

    Does your solution account for all four points of intersection?

    Edit: Never mind. I missed the part that said that R is restricted to the upper half-plane.
    Actually it's good that you pointed that out. I wasn't looking at the lower half plane very carefully. I had simply assumed there were no other crossings and apparently there are. Thanks for the comment.

    -Dan
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