how can you prove that the volume of a sphere is (4/3)*pi*r^3?

Circle center (0,0) , radius = r

x^2 + y^2 = r^2

y^2 = r^2 - x^2

Rotated about the x axis , volume of a sliced disk:

= pi(r^2 - x^2) dx

Volume of the whole sphere:

V = ∫ pi(r^2 - x^2) dx from x = -r to r

= pi[xr^2 - x^3/3] from -r to r

=pi[(r^3 - r^3/3) - (-r^3 + r^3/3)]

= pi[2r^3 -2r^3/3]

V = (4/3) * pi *r^3

In the x,y,z space put a circle cenered at the origin and on on of the planes circle has radisu r. Take the volume of revolution of the circle about one of its plane's axes.

y^2 + x^2 = r^2 or y^2 = r^2-x^2

Volume is then the recvlution integral from -r to r of (F(x)^2*pi)

= pi(r^2*x - x^3/3) -r to r get

pi((r)^3 -r^3/3)) - (-r^3 - r^3/3) = 2/3pir^3 + 2/3piR^2 = 4/3pir^3

you can prove it with and integral of a revolution solid of a circle with equation x²+y²=r²

the volume diferential is

dv= πx²dy

r

∫dv=∫(πx²dy) to find the integration we sustitute x² by r²-y²

-r

r

V=∫(π(r²-y²)dy) =π(r²y-y³/3)=π[(r²(r)-r³/3)-{(r²)(-r)-(-r)³/3}]

-r

=π[2r³/3+2r³/3]

=(4/3)πr³

The proof involves calculus (integration)

See http://www.math.hmc.edu/calculus/tutorials/volume/

But here's an interesting discussion that does not require calculus:

http://mathforum.org/library/drmath/view/66638.htm...

Use analytic geometry and have at it.