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In mathematics —in particular, in multivariable calculus —a volume integral refers to 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.
In coordinates It can also mean a triple integral within a region
D
⊂
R
3
{\displaystyle D\subset \mathbb {R} ^{3}}
function
f
(
x
,
y
,
z
)
,
{\displaystyle f(x,y,z),}
∭
D
f
(
x
,
y
,
z
)
d
x
d
y
d
z
.
{\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz.}
A volume integral in cylindrical coordinates is
∭
D
f
(
ρ
,
φ
,
z
)
ρ
d
ρ
d
φ
d
z
,
{\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho \,d\rho \,d\varphi \,dz,}
and a volume integral in spherical coordinates (using the ISO convention for angles with
φ
{\displaystyle \varphi }
θ
{\displaystyle \theta }
conventions )) has the form
∭
D
f
(
r
,
θ
,
φ
)
r
2
sin
θ
d
r
d
θ
d
φ
.
{\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta \,dr\,d\theta \,d\varphi .}
Example 1 Integrating the equation
f
(
x
,
y
,
z
)
=
1
{\displaystyle f(x,y,z)=1}
∫
0
1
∫
0
1
∫
0
1
1
d
x
d
y
d
z
=
∫
0
1
∫
0
1
(
1
−
0
)
d
y
d
z
=
∫
0
1
(
1
−
0
)
d
z
=
1
−
0
=
1
{\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}1\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}(1-0)\,dy\,dz=\int _{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 density function on the unit cube then the volume integral will give the total mass of the cube. For example for density function:
{
f
:
R
3
→
R
(
x
,
y
,
z
)
⟼
x
+
y
+
z
{\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\(x,y,z)\longmapsto x+y+z\end{cases}}}
the total mass of the cube is:
∫
0
1
∫
0
1
∫
0
1
(
x
+
y
+
z
)
d
x
d
y
d
z
=
∫
0
1
∫
0
1
(
1
2
+
y
+
z
)
d
y
d
z
=
∫
0
1
(
1
+
z
)
d
z
=
3
2
{\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}(x+y+z)\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}\left({\frac {1}{2}}+y+z\right)\,dy\,dz=\int _{0}^{1}(1+z)\,dz={\frac {3}{2}}}
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