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Topics in Calculus

Fundamental theorem
Limits of functions
Continuity
Mean value theorem

Multivariable calculus 

Matrix calculus
Partial derivative
Multiple integral
Line integral
Surface integral
Volume integral
Jacobian

In mathematics — in particular, in multivariable calculus — a volume integral refers to an integral over a 3-dimensional domain.

Volume integral is a triple integral of the constant function 1, which gives the volume of the region D, that is, the integral

\operatorname{Vol}(D)=\iiint\limits_D dx\,dy\,dz.

It can also mean a triple integral within a region D in R3 of a function f(x,y,z), and is usually written as:

\iiint\limits_D f(x,y,z)\,dx\,dy\,dz.

A volume integral in cylindrical coordinates is

\iiint\limits_D f(r,\theta,z)\,r\,dr\,d\theta\,dz,

and a volume integral in spherical coordinates (using the standard convention for angles) has the form

\iiint\limits_D f(r,\theta,\phi)\,r^2 \sin\theta \,dr \,d\theta\, d\phi .

Example

Integrating the function f(x,y,z) = 1 over a unit cube yields the following result:

 \iiint \limits_0^1 1 \,dx\, dy \,dz = \iint \limits_0^1 (1 - 0) \,dy \,dz = \int \limits_0^1 (1 - 0) dz = 1 - 0 = 1

So the volume of the unit cube is 1 as expected. This is rather trivial however and a volume integral is far more powerful. For instance if we have a scalar function \begin{align} f\colon \mathbb{R}^3 &\to \mathbb{R} \end{align} describing the density of the cube at a given point (x,y,z) by f = x + y + z then performing the volume integral will give the total mass of the cube:

 \iiint \limits_0^1 \left(x + y + z\right) \, dx \,dy \,dz = \iint \limits_0^1 \left(\frac 12 + y + z\right) \, dy \,dz = \int \limits_0^1 \left(1 + z\right) \, dz = \frac 32

See also

External links


Topics in Calculus
Fundamental theorem
Limits of functions
Continuity
Mean value theorem
Multivariable calculus 
Matrix calculus
Partial derivative
Multiple integral
Line integral
Surface integral
Volume integral
Jacobian

In mathematics — in particular, in multivariable calculus — a volume integral refers to an integral over a 3-dimensional domain.

Volume integral is a triple integral of the constant function 1, which gives the volume of the region D, that is, the integral

\operatorname{Vol}(D)=\iiint\limits_D dx\,dy\,dz.

It can also mean a triple integral within a region D in R3 of a function f(x,y,z), and is usually written as:

\iiint\limits_D f(x,y,z)\,dx\,dy\,dz.

A volume integral in cylindrical coordinates is

\iiint\limits_D f(r,\theta,z)\,r\,dr\,d\theta\,dz,

and a volume integral in spherical coordinates (using the standard convention for angles) has the form

\iiint\limits_D f(r,\theta,\phi)\,r^2 \sin\theta \,dr \,d\theta\, d\phi .

Example

Integrating the function f(x,y,z) = 1 over a unit cube yields the following result:

\iiint \limits_0^1 1 \,dx\, dy \,dz = \iint \limits_0^1 (1 - 0) \,dy \,dz = \int \limits_0^1 (1 - 0) dz = 1 - 0 = 1

So the volume of the unit cube is 1 as expected. This is rather trivial however and a volume integral is far more powerful. For instance if we have a scalar function \begin{align} f\colon \mathbb{R}^3 &\to \mathbb{R} \end{align} describing the density of the cube at a given point (x,y,z) by f = x+y+z then performing the volume integral will give the total mass of the cube:

\iiint \limits_0^1 \left(x + y + z\right) \, dx \,dy \,dz = \iint \limits_0^1 \left(\frac 12 + y + z\right) \, dy \,dz = \int \limits_0^1 \left(1 + z\right) \, dz = \frac 32

See also

External links








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