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Math Help - (possibly) unsolvable integral

  1. #1
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    (possibly) unsolvable integral

    Ok, so I was bored during AP Calculus after the test on integrals, and decided to entertain myself with a seemingly innocent math problem...

    if you take a semi-circle represented by the equation
    \sqrt[2]{r^2-x^2}
    and divide the semi-circle into fourths, you end up with one-eight of a circle, with an area of
    \frac{\Pi*r^2}{8}
    I want to know when the integral of the equation, represented by
    \frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}} from -R to X is equal to \frac{\Pi*r^2}{8}.
    Basically, I need the following equation solved for X:
    \frac{\Pi*r^2}{8}=\frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}}

    BTW, this problem will get me some SERIOUS extra credit points
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by crazygamer67688 View Post
    Ok, so I was bored during AP Calculus after the test on integrals, and decided to entertain myself with a seemingly innocent math problem...

    if you take a semi-circle represented by the equation
    \sqrt[2]{r^2-x^2}
    and divide the semi-circle into fourths, you end up with one-eight of a circle, with an area of
    \frac{\Pi*r^2}{8}
    I want to know when the integral of the equation, represented by
    \frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}} from -R to X is equal to \frac{\Pi*r^2}{8}.
    Basically, I need the following equation solved for X:
    \frac{\Pi*r^2}{8}=\frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}}

    BTW, this problem will get me some SERIOUS extra credit points
    What I think you need to do is evaluate \int_{-r}^{x}\left[\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\arcsin{\frac{x}{r}}\right]\,dx and see for what x value it equals \frac{\pi r^2}{8}

    Note that \int_{-r}^x\left[\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\arcsin{\frac{x}{r}}\right]\,dx=\tfrac{1}{2}\int_{-r}^x x\sqrt{r^2-x^2}\,dx+\frac{r^2}{2}\int_{-r}^x \arcsin{\frac{x}{r}}\,dx

    For the first integral, make a substitution z=r^2-x^2\implies \,dz=-2x\,dx. Thus you need to deal with -\tfrac{1}{4}\int_{0}^{r^2-x^2} \sqrt{z}\,dz. This yields -\tfrac{1}{6}\left.\left[z^{\frac{3}{2}}\right]\right|_0^{r^2-x^2}=-\tfrac{1}{6}\sqrt{\left(r^2-x^2\right)^3}

    For the first integral, make a substitution, then apply integration by parts:

    z=\frac{x}{r}\implies\,dz=\frac{1}{r}\,dx.

    Thus, we are left to integrate \frac{r^3}{2}\int_{-1}^{\frac{x}{r}}\sin^{-1}z\,dz

    Now, let u=\sin^{-1}z and \,dv=\,dz

    Thus, \,du=\frac{1}{\sqrt{1-z^2}} and v=z

    Thus, \frac{r^3}{2}\int_{-1}^{\frac x r}\sin^{-1}z\,dz=\frac{r^3}{2}\left[\left.\left[z\sin^{-1}z\right]\right|_{-1}^{\frac x r}-\int_{-1}^{\frac x r}\frac{z\,dz}{\sqrt{1-z^2}}\right] =\frac{r^3}{2}\left[\frac{x}{r}\sin^{-1}\left(\frac{x}{r}\right)+\frac{3\pi}{2}+\left.\l  eft[\sqrt{1-z^2}\right]\right|_{-1}^{\frac{x}{r}}\right]=\frac{xr^2}{2}\sin^{-1}\left(\frac{x}{r}\right)+\frac{3\pi r^3}{4}+\frac{r^2}{2}\sqrt{r^2-x^2}.

    Now find x such that -\tfrac{1}{6}\sqrt{\left(r^2-x^2\right)^3}+\frac{xr^2}{2}\sin^{-1}\left(\frac{x}{r}\right)+\frac{3\pi r^3}{4}+\frac{r^2}{2}\sqrt{r^2-x^2}=\frac{\pi r^2}{8}

    After letting Maple try to compute this...it gets me nowhere...so I believe there is no solution...
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  3. #3
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    Quote Originally Posted by crazygamer67688 View Post
    Ok, so I was bored during AP Calculus after the test on integrals, and decided to entertain myself with a seemingly innocent math problem...

    if you take a semi-circle represented by the equation
    \sqrt[2]{r^2-x^2}
    and divide the semi-circle into fourths, you end up with one-eight of a circle, with an area of
    \frac{\Pi*r^2}{8}
    I want to know when the integral of the equation, represented by
    \frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}} from -R to X is equal to \frac{\Pi*r^2}{8}.
    Basically, I need the following equation solved for X:
    \frac{\Pi*r^2}{8}=\frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}}

    BTW, this problem will get me some SERIOUS extra credit points
    I would do this by geometry, not by calculus. In the picture below, you want the shaded area to be one-eighth of the whole disk. The shaded area consists of the sector with angle α, less the right-angled triangle with base r*cos(α) and height r*sin(α). I won't give the details, since you're supposed to be doing this for extra credit, but you should get the equation \alpha - \tfrac\pi4=\sin\alpha\cos\alpha.

    You won't be able to find a neat analytic solution to that equation, but you might be able to see that it is equivalent to \cos\beta = \beta, where \beta = 2\alpha-\tfrac\pi2. That has a numerical solution somewhere near 0.75 radians. To get a better solution, set your calculator to radian measure, enter 0.75, and hit the cos button repeatedly until the display stabilises at the limiting value.

    Once you have a good value for β, you can convert that to a value for α. Finally, the value of x that you want will be r(1–cos(α)).
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  4. #4
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    wow...thanks. I guess i was trying too hard to do it with integrals, and overlooked the fact that it might be solved with geometry...
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