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Area#Area formulas – Size of a two-dimensional surface; Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities
Consider the linear subspace of the n-dimensional Euclidean space R n that is spanned by a collection of linearly independent vectors , …,. To find the volume element of the subspace, it is useful to know the fact from linear algebra that the volume of the parallelepiped spanned by the is the square root of the determinant of the Gramian matrix of the : (), = ….
Regular polygons; Description Figure Second moment of area Comment A filled regular (equiliteral) triangle with a side length of a = = [6] The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.
The given formula is for the plane passing through the center of mass, which coincides with the geometric center of the cylinder. If the xy plane is at the base of the cylinder, i.e. offset by d = h 2 , {\displaystyle d={\frac {h}{2}},} then by the parallel axis theorem the following formula applies:
If the legs have lengths a, b, c, then the trirectangular tetrahedron has the volume [2] =. The altitude h satisfies [3] = + +. The area of the base is given by [4] =. The solid angle at the right-angled vertex, from which the opposite face (the base) subtends an octant, has measure π /2 steradians, one eighth of the surface area of a unit sphere.
The volume of a tetrahedron can be obtained in many ways. It can be given by using the formula of the pyramid's volume: =. where is the base' area and is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apices to the opposite faces are inversely proportional to the areas of ...
The volume can be computed without use of the Gamma function. As is proved below using a vector-calculus double integral in polar coordinates, the volume V of an n-ball of radius R can be expressed recursively in terms of the volume of an (n − 2)-ball, via the interleaved recurrence relation:
as can be seen by multiplying the previous formula by x n+1, to get the volume under the n-simplex as a function of its vertex distance x from the origin, differentiating with respect to x, at = / (where the n-simplex side length is 1), and normalizing by the length / + of the increment, (/ (+), …, / (+)), along the normal vector.