In 3 dimensions, the volume inside a sphere (that is, the volume of the ball) is given by the formula
where r is the radius of the sphere and π is the constant pi. This formula was first derived by Archimedes, who showed that the volume of a sphere is 2/3 that of a circumscribed cylinder. (This assertion follows from Cavalieri's principle.) In modern mathematics, this formula can be derived using integral calculus, e.g. disk integration to sum the volumes of an infinite number of circular disks of infinitesimal thickness stacked centered side by side along thex axis from x = 0 where the disk has radius r (i.e. y = r) to x = r where the disk has radius 0 (i.e. y = 0).
At any given x, the incremental volume (δV) is given by the product of the cross-sectional area of the disk at x and its thickness (δx):
The total volume is the summation of all incremental volumes:
In the limit as δx approaches zero[1] this becomes:
At any given x, a right-angled triangle connects x, y and r to the origin, hence it follows from Pythagorean theorem that:
Thus, substituting y with a function of xgives:
This can now be evaluated:
Therefore the volume of a sphere is:
Alternatively this formula is found using spherical coordinates, with volume element
In higher dimensions, the sphere (or hypersphere) is usually called an n-ball. General recursive formulas exist for deriving the volume of an n-ball.
For most practical uses, the volume of a sphere can be approximated as 52.4% of the volume of an inscribing cube, since . For example, since a cube with edge length 1 m has a volume of 1 m3, a sphere with diameter 1 m has a volume of about 0.524 m3.
Surface area of a sphereThe surface area of a sphere is given by the following formulaThis formula was first derived by Archimedes, a greek, based upon the fact that the projection to the lateral surface of a circumscribing cylinder (i.e. theGall-Peters map projection) is area-preserving. It is also the derivative of the formula for the volume with respect to r because the total volume of a sphere of radius r can be thought of as the summation of the surface area of an infinite number of spherical shells of infinitesimal thickness concentrically stacked inside one another from radius 0 to radius r. At infinitesimal thickness the discrepancy between the inner and outer surface area of any given shell is infinitesimal and the elemental volume at radius r is simply the product of the surface area at radius r and the infinitesimal thickness.
At any given radius r, the incremental volume (δV) is given by the product of the surface area at radius r(A(r)) and the thickness of a shell (δr):
The total volume is the summation of all shell volumes:
In the limit as δr approaches zero[1] this becomes:
Since we have already proved what the volume is, we can substitute V:
Differentiating both sides of this equation with respect to r yields A as a function of r:
Which is generally abbreviated as:
Alternatively, the area element on the sphere is given in spherical coordinates by:
The total area can thus be obtained by integration: +++++++++++++++++
If you don't now anything about calculus & are looking for formulas, then use the following:
V = (4/3) * (pi)*(r3)
A = 4*(pi)*(r2)
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For the question to have any meaning, the volume should be in cubic metres, not metres. The surface area of a sphere of radius r is 4*pi*r*r and its volume is 4/3*pi*r*r*r. Use the second equation to find the value of the radius, r and then use that value in the first equation to calculate the surface area.
0.4 m-1 is the ration of surface area 588m2 to volume 1372m3 for a sphere.
a sphere
Volume = 4/3*pi*radius3 = 900 cubic inches By making the radius the subject of the above equation gives the sphere a radius of 5.989418137 inches. Surface area = 4*pi*5.9894181372 = 450.7950449 square inches
4 pi r 2