如何使用scipy.integrate来获取截断球体的体积？

Question

我正在努力使用scipy.integrate，我使用了tplquad，但是如何使用integrate来获取（截断的）球体的体积？感谢

import scipy
from scipy.integrate import quad, dblquad, tplquad
from math import*
from numpy import *

R = 0.025235 #radius
theta0 = acos(0.023895) #the angle from the edge of truncated plane to the center of
sphere

def f_1(phi,theta,r):
    return r**2*sin(theta)*phi**0
Volume = tplquad(f_1, 0.0,R, lambda y: theta0, lambda y: pi, lambda y,z: 0.0,lambda
y,z: 2*pi)

print Volume

Answer 1

要按角度截断，使用球面坐标系统很方便。假设radius (r)，theta (t)和phi (p)的定义taken from Arkansas TU为：

然后，您可以截断设置限制：r1 r2 t1 t2 p1 p2：

import scipy
from scipy.integrate import quad, dblquad, tplquad
from numpy import *
# limits for radius
r1 = 0.
r2 = 1.
# limits for theta
t1 = 0
t2 = 2*pi
# limits for phi
p1 = 0
p2 = pi

def diff_volume(p,t,r):
    return r**2*sin(p)

volume = tplquad(diff_volume, r1, r2, lambda r:   t1, lambda r:   t2,
                                      lambda r,t: p1, lambda r,t: p2)[0]

要按平面截断，可以方便地使用笛卡尔坐标系(x,y,z)，其中x**2+y**2+z**2=R**2（see mathworld）。在这里，我将截断一半的球体来证明：

from `x1=-R` to `x2=R`<br>
from `y1=0` to `y2=(R**2-x**2)**0.5`<br>
from `z1=-(R**2-x**2-y**2)**0.5` to `z2=(R**2-x**2-y**2)**0.5`<br>
(an useful example using lambdas):

R= 2.
# limits for x
x1 = -R
x2 = R

def diff_volume(z,y,x):
    return 1.

volume = tplquad(diff_volume, x1, x2,
                 lambda x: 0., lambda x: (R**2-x**2)**0.5,
                 lambda x,y: -(R**2-x**2-y**2)**0.5,
                 lambda x,y:  (R**2-x**2-y**2)**0.5 )[0]