用R语言(积分)用3个平面切割的球体的体积

Question

我想要计算球的一部分的体积(V)，这是球与三个球(x=0，y=0和z=1.5)相交的结果。我使用的是R语言，这是我的代码。我尝试了两种不同的方法，使用笛卡尔坐标和极坐标。他们都给出了否定的答案。

##  Compute the Volume between 3 planes x=0, y=0 and z=1.5 and a sphere
library("pracma", lib.loc="~/R/win-library/3.1")
f <- function(x, y)  (sqrt(4 -x^2 - y^2) - 1.5 ) # here the function(x,y)  is subtracted with -1.5 which represents the plane z=1.5 
xmin <- 0
xmax <- 2
ymin <- 0 
ymax <- function(x) (sqrt(4 - x^2))
I <- integral2(f, xmin, xmax, ymin, ymax)
I$Q   # Integral over sector is not so exact

# Exact Volume from AutoCAD V=0.3600



##  Volume of the sphere: use polar coordinates
f0 <- function(x, y) (sqrt(4 - x^2 - y^2)-1.5)  # for x^2 + y^2 <= 4 the f(x,y) means z changes between zmin=1 and zmax= sqrt(4-x^2-y^2)
fp <- function(t, r) r * f0(r*cos(t), r*sin(t))
quad2d(fp, 0, pi/2, 0, 2, n = 101)  # -0.523597

正确的答案是V= 0.3600。谁能给我个提示吗？

干杯

Answer 1

您的X区域集成涵盖了f(x,y)-1.5为负的区域，也涵盖了正的领域.球面与直线z=1.5的交点是半径sqrt(7/4)的圆(使用毕达哥拉斯)，因此适当地调整限制，您可以得到：

library(pracma)
f <- function(x, y)  (sqrt(4 -x^2 - y^2) - 1.5 ) # here the function(x,y)  is subtracted with -1.5 which represents the plane z=1.5 
xmin <- 0
xmax <- sqrt(7/4)
ymin <- 0 
ymax <- function(x) (sqrt(7/4 - x^2))
I <- integral2(f, xmin, xmax, ymin, ymax)
I$Q   # Integral over sector is not so exact
# [1] 0.3599741

非常接近你的预期。

Answer 2

对于r=2的球面和球壳与x=1、y=1、z=1三平面的交点体积，我已经解决了。

##  Compute the Volume between 3 planes x=1.0, y=1.0 and z=1.0 and a sphere
library(pracma)
f <- function(x, y)  (sqrt(4 -x^2 - y^2) - 1 ) # here the function(x,y) is  subtracted with -1.5 which represents the plane z=1.5 
xmin <- 1
xmax <- sqrt(2)
ymin <- 1 
ymax <- function(x) (sqrt(3 - x^2))
I <- integral2(f, xmin, xmax, ymin, ymax)
I$Q   # 
# [1] 0.01520549
# Exact Volume from AutoCAD: 0.0152