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  1. #1
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    tricky integral

    I'm trying to find the arc length of (x^3)(y^2) = 1 from (1,1) to (9,1/27). I tried parameterizing the function as r(t) = (t,t^-3/2) with no luck. I can't seem to find a way to evaluate the integral using u-substitution or trig sub. Any ideas? Thanks.
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    Re: tricky integral

    You don't need parametrization. You can write it as 2 functions: y=x^{-\frac32} and y=-x^{-\frac32}.
    Note that this function is same when reflected about x-axis, so you can find arc length of y=x^{-\frac32} and then just double result.
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    Re: tricky integral

    Quote Originally Posted by simamura View Post
    You don't need parametrization. You can write it as 2 functions: y=x^{-\frac32} and y=-x^{-\frac32}.
    Note that this function is same when reflected about x-axis, so you can find arc length of y=x^{-\frac32} and then just double result.
    It looks like the original poster is only trying to find the length of a portion of the upper arc. Notice that he wanted the arc length between the points (1,1) and \left(9,\frac1{27}\right).

    Finding the arc length of y=x^{-3/2} leads to the integral \int_1^9\sqrt{1+\frac9{4x^5}}\,dx, which cannot be evaluated in terms of elementary functions. Hence the poster's question.

    We can find a numeric solution, however. The length appears to be approximately 8.2373.
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    Re: tricky integral

    You're right. Integral can be only calculated approximately.
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    Re: tricky integral

    Thanks for confirming my suspicions. I think perhaps the instructor made a typo on the assignment. I tried all sorts of crazy stuff like a trig substitution that led to a integration by parts. It just didn't seem like it was leading anywhere, except maybe driving me crazy lol.
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