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Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. [1]
If f is the characteristic function of the diagonal of X×Y, then integrating f along X gives the 0 function on Y, but integrating f along Y gives the function 1 on X. So, the two iterated integrals are different. This shows that Tonelli's theorem can fail for spaces that are not σ-finite no matter which product measure is chosen.
The same technique can be used to give an inductive proof of the volume formula. The base cases of the induction are the 0-ball and the 1-ball, which can be checked directly using the facts Γ(1) = 1 and Γ( 3 / 2 ) = 1 / 2 · Γ( 1 / 2 ) = √ π / 2 .
In Cartesian coordinates, the divergence of a continuously differentiable vector field = + + is the scalar-valued function: = = (, , ) (, , ) = + +.. As the name implies, the divergence is a (local) measure of the degree to which vectors in the field diverge.
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 ...
A surface integral generalizes double integrals to integration over a surface (which may be a curved set in space); it can be thought of as the double integral analog of the line integral. The function to be integrated may be a scalar field or a vector field. The value of the surface integral is the sum of the field at all points on the surface.
The main idea is to express an integral involving an integer parameter (e.g. power) of a function, represented by I n, in terms of an integral that involves a lower value of the parameter (lower power) of that function, for example I n-1 or I n-2. This makes the reduction formula a type of recurrence relation. In other words, the reduction ...
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.