21112007 Integrals ROCK!
Given:

A = πR²
R² = x² + y²
∴ y(x) = (R²  x²)1/2
V = 

π ⋅ y(x)2 dx 
∴V = 

π ((R²  x²)1/2 )2 dx 
V = π 

(R²  x²) dx 
V = π 

R² dx    π 

x² dx 
V = πR2 

dx    π 

x2 dx 
V = πR2  (  x  )  R

  π  (  x3
3

)  R

V = πR2  (  (R(R)  )    π
3

(  R3 (R3 )  ) 
V = πR3 + πR3  πR3 /3  πR3 /3
V = (2  2/3)πR3 
V =  4
3

πR3 
Not long ago, maybe a few weeks, I thought I'd refresh my memory on the trifle of calculus I learned in my OAC classes in high school. I started by going out and buying a new book on the subject, and I must say, I'm very happy with the purchase I made. The book is The Humongous Book of Calculus Problems, by W. Michael Kelley. It has 1000 questions in it, which get progressively more challenging throughout the book. With each question, he walks you through a stepbystep procedure of solving it, and has a battery of side notes to clarify things. It's a great book to learn from, and it's what I've learned from this book that inspired me to write this post today.
I remember a time when I was quite young, still in grade school, and learning some basic geomety. It is a specific moment that I remember, when I wondered if I could figure out the volume of a sphere based on the the equations I knew for a circle. I didn't of course. For a kid that age though, I'd say I had a pretty good grasp of things. My idea at the time involved taking the area of a circle, and multiplying it by half the circumference of that same circle (the logic there I won't bother to explain). Of course, it didn't work, and I later learned that the volume of a sphere is 4/3πr³.
Although that equation is of course correct, I was rather disappointed when I learned it. I did not understand why that equation worked, but merely memorized it for my later convenience. That stewed in the back of my brain for years, until I hit a question in the aforementioned book which I was very happy to solve. That question being exactly what I wanted to know then: given that the area of a circle is πr², show that the volume of a sphere is 4/3πr³
The way to solve it is like this: Take the basic equation for a circle:
x² + y² = r²
Now rearrange that equation, solving it for y:
y² = r²  x²
y = √(r²  x²)
This gives you the ycoordinate(s) of any point on a circle, given it's x coordinate
and radius.
The trick here is to envision that circle as a projection of a sphere. If you look at the circle in 2D on a piece of paper, you can envision it as being merely a center slice of a 3D sphere. Now imagine that sphere divided into slices that stick out of the paper toward you. In that case, that same equation represents the radius of those slices.
Slicing things is of course one of the fundamental parts of calculus. It's by using integrals that we can effectively take an infinite number of slices, and add them all together, giving us the volume of the sphere.
In the case of a sphere, all of those slices are circles, their area being πr². Their radius will change depending on what part of the sphere we're slicing them from. As mentioned above though, that radius will be √(r²  x²). With this information, we can work out the volume of the sphere as a whole. The steps in doing so are what you see on the right.
I have to say, being able to work that out on my own steam really made my day.
Now, the coolest thing about calculus is that you can take one equation, and transform
it to give you other equations. If you take that same equation that we have for the
volume of a sphere, we can actually take it's derivative to get the surface area:
V = 4/3πr³
∴ V' = 4πr²
∴ A = 4πr²
I'm really pleased to be able to gain a good understanding of why those equations are as
they are.