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Integral as area between two curves. Double integral as volume under a surface z = 10 − ( x 2 − y 2 / 8 ).The rectangular region at the bottom of the body is the domain of integration, while the surface is the graph of the two-variable function to be integrated.
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:
In other words, a volume form gives rise to a measure with respect to which functions can be integrated by the appropriate Lebesgue integral. The absolute value of a volume form is a volume element, which is also known variously as a twisted volume form or pseudo-volume form. It also defines a measure, but exists on any differentiable manifold ...
This double integral can be defined using Riemann sums, and represents the (signed) volume under the graph of z = f(x,y) over the domain R. [40] Under suitable conditions (e.g., if f is continuous), Fubini's theorem states that this integral can be expressed as an equivalent iterated integral [41]
It can be thought of as the double integral analogue of the line integral. Given a surface, one may integrate over this surface a scalar field (that is, a function of position which returns a scalar as a value), or a vector field (that is, a function which returns a vector as value).
However, if we give X×Y the product measure such that the measure of a set is the sum of the Lebesgue measures of its horizontal sections, then the double integral of |f| is zero, but the two iterated integrals still have different values. This gives an example of a product measure where Fubini's theorem fails.
Surface–volume integrals [ edit ] In the following surface–volume integral theorems, V denotes a three-dimensional volume with a corresponding two-dimensional boundary S = ∂ V (a closed surface ):