# Math Help - (possibly) unsolvable integral

1. ## (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

2. Originally Posted by crazygamer67688
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...

3. Originally Posted by crazygamer67688
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)).

4. 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...