1. Diagram the symmetry of triple integrals
For details about triple integrals, see: Triple Integral.
The symmetry principle of triple integrals is similar to that of double integrals. For details about the symmetry of double integrals, see: Illustrating the symmetry of double integrals.
Integrade f ( x , y , z ) f(x,y,z)f(x,y,z ) can have different physical meanings. This article takes volume density as an example, which is easy to visualize in three-dimensional space. f ( x , y , z ) f(x,y,z)f(x,y,z ) represents point( x , y , z ) (x,y,z)(x,y,The density at z ) .In fact, the density can be represented by color depth in the integral area in the figure. Drawing the figure is a bit cumbersome, so the integrand is not visualized in the figure . In the figure in this article, only two colors are used to distinguish Ω 1 and Ω 2 \Omega_1, \Omega_2Oh1、Oh2
Can density and mass be negative numbers?For details on this issue, see: [Physics] Can density be a negative number? has no meaning? ——Liquid density reference system
1.1 Integration area Ω \OmegaΩ is aboutxxeven function of x (i.e. with respect to yoz yozyoz plane symmetry)
1.2 Integration area Ω \OmegaΩ is aboutyyEven function of y (i.e. xoz with respect toxozx oz plane symmetry)
1.3 Integration area Ω \OmegaΩ is aboutzzEven function of z (i.e. xoy with respect toxoyx oy plane symmetry)
2. Rotation symmetry of triple integrals
Rotation symmetry means that the integration region Ω \OmegaThe expression of Ω is in x, y, zx, y, zThe form remains unchanged after x , y , z
are interchanged, that is, the integral has nothing to do with the integral variable . Example: SupposeΩ = { (x, y, z) ∣ x 2 + y 2 + z 2 ≤ 1 } \Omega=\{( x,y,z)|x^2+y^2+z^2\leq1\}Oh={(x,y,z)∣x2+y2+z2≤1 }，求∭ Ω z 2 dxdydz = \iiint_{\Omega}z^2dxdydz=∭Ohz2dxdydz=
integration areaΩ \OmegaΩ expression:x 2 + y 2 + z 2 ≤ 1 x^2+y^2+z^2\leq1x2+y2+z2≤1.
Based on the original expression, variablesx, yx, yAfter exchanging x and y, the expression is: y 2 + x 2 + z 2 ≤ 1 y^2+x^2+z^2\leq1y2+x2+z2≤1 , the expression remains unchanged.
Based on the original expression, the variablesy, zy, zThe expression after swapping y and z is: x 2 + z 2 + y 2 ≤ 1 x^2+z^2+y^2\leq1x2+z2+y2≤1 , the expression remains unchanged.x, zx, z
are based on the original expression.The expression after exchanging x and z is: z 2 + y 2 + x 2 ≤ 1 z^2+y^2+x^2\leq1z2+y2+x2≤1 , the expression remains unchanged.
After verification, the integral region has rotation symmetry, thenzzReplace z with xxx和yyy后 integral size unchanged∭
Ω z 2 dxdydz = ∭ Ω x 2 dxdydz = ∭ Ω y 2 dxdydz \iiint_{\Omega}z^2dxdydz=\iiint_{\Omega}x^2dxdydz=\iiint_{\Omega}y ^2dxdydz∭Ohz2dxdydz=∭Ohx2dxdydz=∭Ohy2 dxdydz
∭ Ω z 2 dxdydz = 1 3 ( ∭ Ω x 2 dxdydz + ∭ Ω y 2 dxdydz + ∭ Ω z 2 dxdydz ) ∭ Ω z 2 dxdydz = 1 3 ( ∭ Ω x 2 + y 2 + z 2 dxdydz ) \iiint_{\Omega}z^2dxdydz=\frac{1}{3}\big(\iiint_{\Omega}x^2dxdydz+\iiint_{\Omega}y^2dxdydz+\iiint_{\Omega} z^2dxdydz\big)\\ ~\\ \iiint_{\Omega}z^2dxdydz=\frac{1}{3}\big(\iiint_{\Omega}x^2+y^2+z^2dxdydz\ big)∭Ohz2dxdydz=31(∭Ohx2dxdydz+∭Ohy2dxdydz+∭Ohz2dxdydz) ∭Ohz2dxdydz=31(∭Ohx2+y2+z2 dxdydz)
Since the integration area is a sphere, it is simpler to use the spherical coordinate system to solve
∭ Ω x 2 + y 2 + z 2 dxdydz = ∫ 0 2 π d θ ∫ 0 π d ϕ ∫ 0 R r 2 ⋅ r 2 sin ϕ dr = 4 5 π R 5 \iiint_{\Omega}x^2+y^2+z^2dxdydz=\int_{0}^{2\pi}d\theta\int_{0}^ {\pi}d\phi\int_{0}^{R}r^2\cdot r^2sin\phi dr=\frac{4}{5}\pi R^5∭Ohx2+y2+z2dxdydz=∫02 p.md i∫0pdϕ∫0Rr2⋅r2sinϕdr=54πRDetermine the 5th
dimension:
∭ Ω z 2 dxdydz = 1 3 ⋅ 4 5 π R 5 = R=1 4 15 π \iiint_{\Omega}z^2dxdydz=\frac{1}{3}\cdot\frac{4 }{5}\pi R^5\translated{\text{R=1}}{=}\frac{4}{15}\pi∭Ohz2dxdydz=31⋅54πR5=R=1154Pi