V = 4/3πr3

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

Where V is the volume of the sphere, r is the radius of the sphere, and Pi is a constant.


The volume of a hemisphere with radius r is the same as the volume of a cylinder with radius and height r, with a right cone of radius and height r with the vertex at the center of the base removed (the vertex is down-- think ice cream cone). To prove this, we will find that the cross-sectional areas of the hemisphere and of the "cylinder less a cone" are equal at any height h (Cavalieri's Principle).

The Hemisphere

Any cross-section of a hemisphere parallel to the great circle will also be a circle; let's assign it a radius of x. If that circle is height h above the base, then it can also be described as part of a right triangle whose legs are h and x, and whose hypotenuse is r, the radius of the hemisphere. We can then describe x with the Pythagorean theorem: x² + h² = r². This can be rewritten as x = sqrt(r²-h²). So the area of the cross-section is π(sqrt(r²-h²)²), which is equal to π(r²-h²). Remember that.

The "Cylinder Less Cone"

The shape of a cross-section of the cylinder less cone is a little trickier, but it's easy to figure out: the area of the base of the cylinder, minus a cross-section of the cone at height h-- we'll give this cross-section a radius of y. So the cross-sectional area is (πr²-πy²). But look at this:

|\    |    /|
| \_y_|   / |
|  \  |h /  |
|  l\ | /   |

The triangle formed by h, y, and the line segment between where y meets the slant height of the cone and the center of the base (l in the diagram) is a 45-45-90 isosceles triangle. Therefore, h = y. So our previous equationr²-πy²) can be replaced with (πr²-πh²), which (when factored) comes out to π(r²-h²)-- the same as the area of a cross-section of the hemisphere at the same height! Therefore, the two solids are of the same volume.

Almost There

We now have to find the volume of the cylinder less cone.

Vcyl = πr²h = πr³
Vcone = (1/3)πr²h = (1/3)πr³
Vtotal = Vcyl-Vcone = (2/3)πr³

But remember, this is the volume of a hemisphere. Double it, and we get, at long last, the volume of a sphere:



Calculus derivation of the volume of a sphere:

I once had a math test for which I needed this formula. I didn't have it memorized so I tried to derive it. I eventually came up with the correct equation but when I looked at my work later I realized I had a bunch of mistakes that just happened to cancel each other out. I've reworked the derivation so that it makes sense. Here it is.

Let A be the region bounded by:
the semi-circle y = |sqrt(r^2 - x^2)|,
the line y = 0,
the line x = -r,
and the line x = r
where r is a constant.

Let V be the volume of the solid defined by rotating A about the x-axis. We'll assume (since I haven't figured out how to prove it) that this solid is spherical.

A slice of the solid taken parallel to the y-axis is a cylinder with volume = pi*y^2*dx. Now, take the sum of the slices from x = -r to r, as dx approaches 0.

pi*y^2 = pi(r^2 - x^2)
The Indefinite Integral of 'pi(r^2 - x^2)dx' is 'pi(xr^2 - 1/3x^3) + c'
And the Definite Integral from x = -r to x = r gives us:

V = pi(r^3 - 1/3r^3 + c - (-r^3 + 1/3r^3 + c)) (by the Fundamental Theorem of Calculus)
= pi(2/3r^3 + r^3 - 1/3r^3)
= pi(4/3r^3)

which is the familiar formula: '4/3pi*r^3'.

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