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My curiosity was piqued by another poster who's trying to find the area of a lune using calculus.

I wanted to do this now. So I don't highjack his thread, I'm making a new post about it.

First, the geometric picture of a lune that constitutes my basic plan of attack. First I want to solve a special case of a lune, where the larger circle intersects the smaller circle at the smaller circle's widest point. (However, I can't even get there. I can't find the area of a quarter or semi circle by integration. So as you will see, I'm going to end this post by asking for help with this easier problem).

Consider the smaller circle, of radius ##r##, as being centered at the origin, and bigger circle, of radius R, offset along the y axis by -b.

The two equations which express the height ##y## of these circles as a function of x in the first quadrant are:

##

f(x) = \sqrt{r^{2}-x^{2}} \\

g(x) = \sqrt{R^{2}-x^{2}} - b \\

##

Now the area of the smaller circle above the x axis should be equal to

##A_{f} = 2\int_{0}^{r}f(x)dx##

I should be able to calculate this if I can find an antiderivative for F. This turns out to be hard.

##f(x) = \sqrt{r^{2}-x^{2}} \\

= \sqrt{r^{2}(1-\frac{x^2}{r^2}} \\

= r\cdot \sqrt{1-(\frac{x}{r})^{2}} \\

##

Then after having a heck of a time trying to find an antiderivative for that, I realized that I need to fish for something "pi-ish".

Recognizing that ##r \cos \theta = x## in the first quadrant, we have that ##\cos \theta = \frac{x}{r}##

##

f(\theta) = r \cdot \sqrt{1- \cos ^{2}\theta} \\

= r \cdot \sin \theta \\

##

And in retrospect, I could have just started there in polar coordinates.

Now, considering the first quandrant in a clockwise sense, we start when

##cos \theta = \frac{0}{r}## and stop when ##\cos \theta = \frac{r}{r} = 1 ##. So, in radians,

##cos^{-1}(0) = \frac{\pi}{2} \\

cos^{-1}{1} = 0##

So we integrate from ##\frac{\pi}{2}## to ##0##.

##A = -\int_{0}^{\frac{\pi}{2}} r \sin \theta d\theta##

So now, it seems this should be easy: i just need an antiderivative for ##r \sin \theta##. I think it's ##-r \cos \theta## by the rule for constants.

The answer for a quarter circle should be ## -r( -\cos 0 + \cos \frac{pi}{2} ) = r##

But this answer can't be right, because it doesn't include pi. How am I getting off track here?

I wanted to do this now. So I don't highjack his thread, I'm making a new post about it.

First, the geometric picture of a lune that constitutes my basic plan of attack. First I want to solve a special case of a lune, where the larger circle intersects the smaller circle at the smaller circle's widest point. (However, I can't even get there. I can't find the area of a quarter or semi circle by integration. So as you will see, I'm going to end this post by asking for help with this easier problem).

Consider the smaller circle, of radius ##r##, as being centered at the origin, and bigger circle, of radius R, offset along the y axis by -b.

The two equations which express the height ##y## of these circles as a function of x in the first quadrant are:

##

f(x) = \sqrt{r^{2}-x^{2}} \\

g(x) = \sqrt{R^{2}-x^{2}} - b \\

##

Now the area of the smaller circle above the x axis should be equal to

##A_{f} = 2\int_{0}^{r}f(x)dx##

I should be able to calculate this if I can find an antiderivative for F. This turns out to be hard.

##f(x) = \sqrt{r^{2}-x^{2}} \\

= \sqrt{r^{2}(1-\frac{x^2}{r^2}} \\

= r\cdot \sqrt{1-(\frac{x}{r})^{2}} \\

##

Then after having a heck of a time trying to find an antiderivative for that, I realized that I need to fish for something "pi-ish".

Recognizing that ##r \cos \theta = x## in the first quadrant, we have that ##\cos \theta = \frac{x}{r}##

##

f(\theta) = r \cdot \sqrt{1- \cos ^{2}\theta} \\

= r \cdot \sin \theta \\

##

And in retrospect, I could have just started there in polar coordinates.

Now, considering the first quandrant in a clockwise sense, we start when

##cos \theta = \frac{0}{r}## and stop when ##\cos \theta = \frac{r}{r} = 1 ##. So, in radians,

##cos^{-1}(0) = \frac{\pi}{2} \\

cos^{-1}{1} = 0##

So we integrate from ##\frac{\pi}{2}## to ##0##.

##A = -\int_{0}^{\frac{\pi}{2}} r \sin \theta d\theta##

So now, it seems this should be easy: i just need an antiderivative for ##r \sin \theta##. I think it's ##-r \cos \theta## by the rule for constants.

The answer for a quarter circle should be ## -r( -\cos 0 + \cos \frac{pi}{2} ) = r##

But this answer can't be right, because it doesn't include pi. How am I getting off track here?

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