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Differentiating and Integrating the Formula for Area of Circle


Date: 05/11/98 at 23:22:53
From: Joey Bohanon
Subject: Derivatives with circles

I have recently been reading up on calculus in a book my dad got me. I 
noticed that if f = pi*r^2 (the area of a circle), then f' = 2*pi*r 
(the circumference of a circle). I also integrated pi*r^2 and came up 
with (pi*r^3)/3. Is the result of this integral a formula for some 
sort of solid? If so, could this be used in figuring out volumes (I'm 
not sure what they're called) of 4-dimensional objects, and so on? 
What is the significance of the circumference being the derivative of 
the area?

Thanks a bunch,
Joey


Date: 05/14/98 at 11:39:27
From: Doctor Schwa
Subject: Re: Derivatives with circles

Joey,
I'll take your third question first. It is such a good question, I 
thought for sure someone would have asked it before, but when I 
searched the Dr. Math archives for "circumeference," "area," and 
"derivative," I didn't find anything. So it's up to me to meet the 
challenge.

The derivative is the rate of change. If I let A = pi*r^2, then what 
does dA/dr mean? It's the limit of:

    change in area
   ----------------
   change in radius

So your question becomes: what happens to the area when the radius is 
increased a tiny amount?

The change in area is the area of a thin washer, of thickness dr in 
the radius direction. If you imagine cutting that washer and unbending 
it to make a straight line, you'd have (approximately) a rectangle, 
with a thickness of dr and a length equal to the circumference of 
the circle.

Thus dA = (circumference) * dr.

This is similar in some ways to the 2pi*r*h formula for the area of 
the side (lateral area) of a cylinder, with the cutting and unrolling 
idea.

You also ask about the integral of pi*r^2, which is equal to 
(pi*r^3)/3. First, this integral is a formula for some sort of solid, 
but not quite in the same way as the relationship between area and 
circumference. This one I would think of as adding up the areas of a 
pile of circles, increasing in radius -- that is, a cone. If you want 
more explanation of why this is a good way to visualize it, let me 
know. But indeed, (pi*r^3)/3 is the volume of a cone of radius r and 
height r.

The derivative of the volume of a sphere is its surface area, and yes, 
the same sort of thing holds in higher dimensions. So if you wanted 
the "volume" of a four-dimensional "cone" made by stacking up spheres
in the fourth dimension (in the same way as a 3D cone is made by
stacking up circles), you could integrate (4/3)*pi*r^3*dr.

And if you knew the formula for the "hypervolume" of a 4D sphere, you 
could find it by integrating the "hyper surface area" formula, if only 
you had that. Or you could find the "surface area" if you only knew 
the volume.

But actually, you can find the "volume" of a 4D "hypersphere." Think 
about how to use integrals to find the area of a circle, or the volume 
of a sphere, and then try to extend the pattern. Let me know how it 
goes!

-Doctor Schwa, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Conic Sections/Circles
High School Geometry
High School Higher-Dimensional Geometry

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