Volume integral

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 function.

In coordinates

Often the volume integral is represented in terms of a differential volume element $dV=dxdydz$. $$\underset{D}{\iiint }f(x,y,z)dV.$$ It can also mean a triple integral within a region $D\subset {\mathbb{ℝ}}^3$ of a function $f(x,y,z),$ and is usually written as: $$\underset{D}{\iiint }f(x,y,z)dxdydz.$$ A volume integral in cylindrical coordinates is $$\underset{D}{\iiint }f(\rho ,\phi ,z)\rho d\rho d\phi dz,$$ and a volume integral in spherical coordinates (using the ISO convention for angles with $\phi$ as the azimuth and $\theta$ measured from the polar axis (see more on conventions)) has the form $$\underset{D}{\iiint }f(r,\theta ,\phi )r^2\sin \theta drd\theta d\phi .$$ The triple integral can be transformed from Cartesian coordinates to any arbitrary coordinate system using the Jacobian matrix and determinant. Suppose we have a transformation of coordinates from $(x,y,z)\mapsto (u,v,w)$. We can represent the integral as the following. $$\underset{D}{\iiint }f(x,y,z)dxdydz=\underset{D}{\iiint }f(u,v,w)\left|\frac{\partial (x,y,z)}{\partial (u,v,w)}\right|dudvdw$$ Where we define the Jacobian determinant to be. $$\mathbf{𝐉}=\frac{\partial (x,y,z)}{\partial (u,v,w)}=\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\\ \end{array}\right|$$

Example

Integrating the equation $f(x,y,z)=1$ over a unit cube yields the following result: $$\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}1dxdydz=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}(1-0)dydz=\underset{0}{\overset{1}{\int }}(1-0)dz=1-0=1$$

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 density function on the unit cube then the volume integral will give the total mass of the cube. For example for density function: $$\{\begin{array}{l}f:{\mathbb{ℝ}}^3\to \mathbb{ℝ}\\ f:(x,y,z)\mapsto x+y+z\end{array}$$ the total mass of the cube is: $$\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}(x+y+z)dxdydz=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}(\frac{1}{2}+y+z)dydz=\underset{0}{\overset{1}{\int }}(1+z)dz=\frac{3}{2}$$

See also

• Divergence theorem
• Surface integral
• Volume element
• Line element
• Line integral

External links

• Template:Springer