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 Area-Volume Formulas for N-Dimensional Ellipsoids

In two dimensions there is the formula that the area of an elliptical disk enclosed withing an ellipse of semi-axes of a and b; i.e., area = πab. In three dimensions the formula for the volume enclosed within an ellipsoid with semi-radii of of a, b and c is (4/3)πabc.

The area of an elliptical disk can be found as the limit of a sequence of approximations in which the disk is covered by a set of rectangles as shown in the diagram below.

In the above construction the vertical axis of the disk is divided into m equal intervals. The width of a rectangle is the width of the disk at that height. As the subdivision of the vertical axis of the disk becomes finer and finer the sum of the areas of the rectangles approaches a limit which is called the area of the disk.

The process can be represented algebraically by noting that the equation of the ellipse which encloses the disk can be expressed in terms of the horizontal and vertical distances from the center, x and y, as (x/a)2 + (y/b)2 = 1. Thus at a height y the width of the disk is 2r where r = a(1-(y/b)2)1/2. The limit can be expressed as an integral; i.e.,

#### V2(a,b) = ∫b-b2a(1-(y/b)2)1/2dy

By symmetry this can be reduced to:

#### V2(a,b) = 2∫b02a(1-(y/b)2)1/2dy

By a change of variable to z=y/b (and thus dy=bdz) the area reduces to

#### V2(a,b) = 2∫102a(1-z2)1/2bdz = 4ab∫10(1-z2)1/2dz

Since ∫(1-z2)1/2dz = (z/2)(1-z2)1/2 + (1/2)sin-1(z) the definite integral reduces to (1/2)π/2 and hence the area of the elliptical disk reduces to

#### V2(a,b) = 4ab[(1/2)π/2] = πab

For the volume enclosed within an ellipsoid of semi-radii a,b and c the procedure is similar. The equation for the ellipsoid is (x/a)2+(y/b)2+(z/c)2=1. Thus at a level z the cross-section is an elliptical disk enclosed within an ellipse with semi-radii of a(1-(z/c)2)1/2 and b(1-(z/c)2)1/2. The volume of the ellipsoid can be expressed as

#### V3(a,b,c) = 2∫c0V2(a(1-(z/c)2)1/2b(1-(z/c)2)1/2)dz = 4∫c0πab(1-(z/c)2)dz

With a change of variable to s=z/c the formula becomes

#### V3(a,b,c) = 2πabc∫10(1-s2)ds = 2πabc = 2πabc[s - (1/3)s3]10 = 2πabc[1-(1/3)] = 2πabc[2/3] = (4/3)πabc

For 4 dimensions the formula reduces to

#### V4(a,b,c,d) = (8/3)πabcd∫10(1-s2)3/2ds

The definite integral reduces to (3/8)(π/2) so the value of V4(a,b,c,d) is equal to (1/2)π2abcd.