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In mathematics (particularly multivariable calculus), a volume integral (∭) is an integral over a 3-dimensional domain; that is, it is a special case of multiple integrals. Volume integrals are especially important in physics for many applications, for example, to calculate flux densities, or to calculate mass from a corresponding density ...
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:
An even larger, multivolume table is the Integrals and Series by Prudnikov, Brychkov, and Marichev (with volumes 1–3 listing integrals and series of elementary and special functions, volume 4–5 are tables of Laplace transforms).
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx.
Much more work is needed to find the volume if we use disc integration. First, we would need to solve y = 8 ( x − 1 ) 2 ( x − 2 ) 2 {\displaystyle y=8(x-1)^{2}(x-2)^{2}} for x . Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow.
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 : (), = ….
Richard Feynman observed that: [1] = [+ ()] which is valid for any complex numbers A and B as long as 0 is not contained in the line segment connecting A and B. The formula helps to evaluate integrals like:
The obvious solution is then to split that surface into several pieces, calculate the surface integral on each piece, and then add them all up. This is indeed how things work, but when integrating vector fields, one needs to again be careful how to choose the normal-pointing vector for each piece of the surface, so that when the pieces are put ...