Finding the area of a circle by Integration
Posted by iamsuhasm on April 14, 2009
Every calculus tutorial that i have read , shows the derivation of πr^2 by either adding together tiny rings or by cutting up the circle into triangles and integrating. So , i thought why not derive the formula , by actually integrating the equation of a circle? And voila to my surprise , it actually worked (I’m still a calculus noob)
So , first we need the function for a circle. Every point on the circle is equidistant from the origin. Hence the equation is clearly x^2 + y^2 = r^2 , where r is the required radius.
Rearranging to get a proper function ,
Now , if i take this function and plot it , it should ideally give me a circle. And it does. Unfortunately , for reasons unknown , Microsoft math is seeming to clip off the negative half of the circle. I dont know why. But that really doesnt matter as i will only be taking the part of the circle in the 1st quarant and will be finding its area. Then we can multiply the area by 4 to get the area of the whole circle.
Now since i have my function , i can find the area of the circle by integrating the function between o and r. This integration will give me the area of a quarter of a circle. We can multiply that area by 4 to get the full area.
Ok. To integrate , we need to find the antiderivative of the circle function. Microsoft math tells me that it is :
The Anti derivative
Now , let me call this function F(x).
Finding F(r) – F(0) should give me the area.
F(0) is very clearly 0 , as both the terms in the equation become zero.
F(r) is
as arctan of (x/sqrt(0)) is apparently 90 = π/2 radians. (I dont know why. Maybe by taking the limit of the function?)
and there you have it folks , ( πr^2 /2) , which is (πr^2/4) !!
Which is precisely the area of a quarter of a circle.
So , now if we multiply it by 4 , we get the area of the whole circle = πr^2 !!!
Isaac said
It’s heartening to see the next generation of calculus students experimenting with math like this, verifying old formulas from the calculus point of view.
I noticed that you noticed that the bottom half of the circle is being clipped off. Actually, you shouldn’t be surprised. By now you probably know why, but in case you don’t, bear in mind that there are TWO square roots for each number: the positive and the negative root. You’ve cut off the negative half of the graph yourself by not including the negative root. But as you cleverly note, it doesn’t matter.
As for the antiderivative, it seems that computer math systems give crappy answers. It’s correct, but it’s in a form that would give you problems. The limit as the arctan function goes to infinity is indeed π/2, but you shouldn’t need that. If you look at any integral table they’ll list the antiderivative you were looking for, in a different form. Instead of arctan(blahblahblah), they have arcsin(x/r). Thus you don’t have to take limits – arcsin(1) is just π/2. (In case you were wondering: yes, the two forms are equivalent. See if you can’t prove it.)
If you’re interested in how to actually do this integral by hand, look in a good textbook for a section about trigonometric substitutions. If you substitute x = r cos t, making sure to also substitute dx = r sin t dt and to change the limits on the integral, your work becomes much easier. When I first learned to do integrals using that trick I was amazed…and had a lot of fun.
Or, you could do the whole problem in polar coordinates. Then it boils down to integrating 1 dx.
Brian said
You don’t seem to have noticed this, but Microsoft Math didn’t literally take the antiderivative. It did the ‘cutting up the circle into triangles and integrating’, and since that left a triangle of error (since that’s polar and the integral is along the x-axis), it added in a correcting factor.
You see, the [(1/2) theta * r^2] is right there – as [(1/2) arctan (x/y) * r^2]. And the [(1/2) x * y] is the correction.
The literal antiderivative is infinitely recursive. What Microsoft Math gave you is considerably easier to work with.
Ian said
If you want a fun computer programming project try and calculate the area the value of Pi using a monte carlo process. The basics of this are to drop as many points as you want into a square by giving it random x,y coordinates with side length of 2R and then from the same data set find a subset of points that are within a circle of radius R. You now have the area of the square of (2R)^2 and the area of the circle Pi*R^2. The ratio of points inside the circle to points inside the square should then be Pi/4. If there are N points inside the square there should be N*Pi/4 points in the circle, then just solve for Pi. Experiment with larger and larger numbers for N to get better and better accuracy. As a physicist I use monte carlo methods for much of my work but I started with this exact problem while in my undergraduate.
I found an example of this http://www.eveandersson.com/pi/monte-carlo-circle