# Proof of Volume of a Sphere

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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

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.

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

Yea, I know I'm slow. Anyway, here's the volume using a triple integral. And I didn't know either what Daniel meant about the volume being zero, and in fact it took me a while to figure it out even after Cepheid explained it.

In spherical coordinates, the problem can be defined as follows:

$$vol=8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^r \rho^2 \sin(\phi)d\rho d\theta d\phi$$

Beautiful isn't it!

So:

$$8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2} \sin(\phi)(\frac{\rho^3}{3}){|}_0^r d\theta d\phi$$

and then:

$$\frac{4r^3 \pi}{3}\int_0^\frac{\pi}{2}sin(\phi)d\phi$$

or:

$$-\frac{4r^3\pi}{3}[0-1]=\frac{4\pi r^3}{3}$$

Don't you just love Calculus!

good.

r is a constant because if you have a sphere, then r is a specific number. For example, a sphere of radius 2, r=2. That's kind of tough to grasp when looking at it for the first time.

Counterpoint gave a triple integral to represent the sphere in the first octant, which is only an eighth of the sphere, which is why he multiplied by eight to get the final volume. His answer was correct, but here is the volume of a sphere in triple integrals without cutting it up:

$$\int_0^{2 \pi }\int_0^{ \pi }\int_0^r\rho^2d\rho d\phi d\theta$$

$$\left( -\cos \phi \Big|_0^{ \pi } \right) (2\pi)\left(\dfrac{r^3}{3}\right)$$

$$\dfrac{4}{3}\pi r^3$$​

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excuse me Jameson there is only one variable in the integration given by you. the "r" is not a variable but is a constant which is actually the limit for the variable "x" (upper limit in this case). To integrate you need to put a trigonometric form of x i.e x=r cos(theta). then choose the limits for "theta" and "theta" becomes the variable instead of "x".

i havent read all the replies, but there is a classic greek proof to this problem.
take a hemisphere of radius "r", a cone and cylinder of radius=height=(r)

then, volume of cylinder is pi r^3 and of the cone is 1/3*pi r^3
thus, all we need is to find the volume of the hemisphere
if 2 solids have the same area of cs fr all arbitary slices, they have equal volumes.
thus we have, the volume sphere =volume of cylinder-vol of cone. thus=2pi/3 r^3
so, volume of a sphere is 4*pi/3 r^3

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BTW,in case you didn't know,the volume of a sphere is ZERO...

Daniel.

Is this because what the OP's really wants to prove is the volume of the closed ball comprising the sphere and its interior?

Forgive what's undoubtedly a simple question, but this isn't my field. I came in because I was interested in the same proof (or, at least, a proof of the same fact).

Also everyone please excuse my name. It seems to have gotten truncated and I'll have to see if the admins will let me change it.

i think one could prove it without calculus.

I haven't read through all this lengthy thread, but I know it can be done; I did it myself many years ago and have been trying to remember the details of the algebra involved. My approach involved imagining a hemisphere with n horizontal disks of equal thickness inscribed therein.

I haven't mastered the tex system yet, either, so I can't lay it all out. Essentially, though, the radius of each slice is one leg of a right triangle of which the other is kr/n where k is that slice's position counting up from the "equator" and n is the number of slices we are using. The Pythagorean theorem gives us the radius of slice k:

sqrt (r^2 - (k^2*r^2)/n^2) )

The volume of each slice is

(pi)*(r/n)*[r^2 - (k^2*r^2)/n^2) ]

so the solution for the whole hemisphere is simply to take the Riemann sum of all the slices. This involves the sum of the first n squares, which is

1/6*n*(n+1)(2n+1)

By decomposing this in the correct way I was able to arrive at a form which converged on the familiar 2/3*(pi)*r^3 value for the volume of the hemisphere, when n is allowed to increase without bound. Now I can't remember the details. I didn't use the cone and cylinder of Archimedes.

I suppose my proof is "calculus-ish", but it's definitely not formal calculus. Certainly there's no triple integral.

i have proved the volume of sphere .it is similar to you .but there is difference. i have placed the sphere on x y z axis.and the origin coincides with center of sphere.if we cut the sphere cross sectionally parallel to y axis then the radius of each cross section y and x will haver following relation
y^2+x^2=r^2
so,y^2=r^2-x^2
and let the distance between each cross be x0. then volume of hemisphere =pi[(r^2-x0^2)x0+(r^2-4x0)x0+(r^2-9x0)x0..............................................r/x0]
=pi[r^3-x0^3(1+4+9.......................................r/x0]
=pi[r^3/-x0^3[{r/x0(r/x0+1)(2r/x0+1)}/6]
=pi[r^3-x0^3[{2r^3/x0^3}/6] because x0 is extremely small
=pi[r^3-r^3/3]
=pi[2r^3/3]
so the volume of sphere will be4/3pi*r^3 double of volume of hemisphere

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let me try i think i have a quite simple way to prove it