Proof of Volume of a Sphere

mathwonk

ok i have actually read more of archimedes and think i know how he found the volume of a sphere, or at least how he proved it. (he discovered it by setting up a lever and balancing the weights of different solids, knowing the centers of gravity of some of them, and deducing that of others.)

basic principles:
1) principle of parallel slices: two solids with equal areas for all plane slices parallel to a given plane, have equal volumes.
2) magnification principle: two pyramids with bases of equal area, have volumes in the same ratio as their heights.

these principles are proved by the method of approximation by blocks or cylinders, since solids with equal plane slices have equal approximating cylinders, and scaling the height merely scales the height of the approximating cylinders. then one proceeds as follows, first for pyramids and cones, then spheres.

step 1) right pyramids of height equal to base edge:
choose 2 opposite vertices on a cube, call them 1 and 2, and join them by a diagonal. choose a face having vertex 2 as a corner, and join every point of this face to vertex 1. this forms a right pyramid. the other two choices of faces having vertex 2 as corner, yield congruent pyramids, by rotation, and all three together make up the cube. thus the given right pyramid has volume 1/3 that of the cube, or 1/3 Bh, where B = area of base, and h = height.

step 2) using magnification principle, one extends the same formula to the case of arbitrary height in comparison to base edge, and using parallel slices one extends the same formula to pyramids which are not "right", but for which the angle to the vertex is arbitrary, since sliding a pyramid over at a new angle does not change the area of parallel slices.

step 3) approximating the base circle by polygons, hence approximating the cone by pyramids, gives the same formula for a cone, V = 1/3 Bh.

step 4) now circumscribe a cylinder about a sphere, and inscribe a double cone (vertex at center, bases at both top and bottom) in the same cylinder. then pythagoras shows that the area of a parallel slice of the cylinder has area equal to the sum of the parallel slices of the sphere and the cone.

Thus the volume of the cylinder equals the sum of the volumes of the cone and the sphere. in particular since the cone has 1/3 the volume of the cylinder, the sphere has 2/3 the volume of the circumscribing cylinder.

And that is how archimedes proved the volume of a sphere.

the by the argument above, viewing the sphere as a limit of pyramids with vertices at the center, he showed the surface area of the sphere, defined as the limit of the areas of the bases of the inscribed pyramids, was 3/R times the volume of the sphere, since tht is the formula for the base area of a pyramid in terms of the volume.

I.e. the volume of a sphere is 1/3 SR where S is the surface area and R is the radius.

and thats that!! hooray for archimedes, who was obviously in almost complete command of the methods of purely integral calculus.

the only thing needing to be added, was the algebraic technique of antidifferentiating the algebraic formula for the area of the parallel slices and getting an algebraic formula for the moving volumes below each slice.

so as far as i know now it had nothing to do with ding up squares of integers at all, quite opposite to my original impression.

mathwonk

moreover archimedes said he could also compute that the volume of a bicylinder, intersection of two perpendicular cylinders, is 2/3 that of a circumscribing cube. his solution of this is lost, but you can guess it if you reflect that the horizontal slices of a bicylinder are intersections of horizontal slices of cylinders, i.e. intersections of rectangles, hence are squares.

thus you want to replace his prior use of a cone by some cone - like figure whose horizontal slices are squares. what do you guess? .........that's right! try it.

so this is archimedes actual work, and this i believe should be taught to every geometry student before attempting calculus. in fact harold jacobs' fine high school geometry book has this calculation of the volume of a sphere near the very end of his book.


i also feel that this use of limits is not properly calculus, but that calculus is the combination of differentiation and integration, found in the fundamental theorem. i.e. i would preserve the term calculus for the use of antidifferentiation to compute the limits archimedes used to define volumes. of course this use of terminology is a matter of preference. note euler also declined to refer to the limits involved in infinite series, as calculus.

arildno

thanks mathwonk!
He becomes greater and greater, the more I get to know his work..

dekyfineboy

When the region between a and b of the function f(x) is rotated about the x-axis, the solid formed will have a volume

(pi)*(integration of f(x)^2). ----------------- 1

so we need the the formula of a circle so that we can put it into the formula

formula of a circle is given by r^2=x^2 + y^2 ---------------- 2
therefore making y the subject y^2=r^2 - x^2 ---------------- 3

put y=f(x) into the equation 1 and the formula for the volume of a sphere will be found.

oscar_osa

You are almost there,

When you integrate (r^2-x^2)dx you will have to fix the limits of integration, meaning, I can recommed from 0 to R, be careful, this is only half of the sphere, when you integrate the result is just r^2.R - R^3/3 this is iqual to R^3 - R^3/3 by just algebra this is equal to 2.R^3/3, as I told you this is just half of the sphere, double this number and you will obtain 4R^3/3, remember that pi was already out of the integration as constant. So to make the story short you have at the end 4.pi.R^3/3

I hope this helps

dekyfineboy

note: the integration will have to have its limits as -r and r since that si the boundries of
the circle

counterpoint

good.