决定把例题用程序都完成一遍。从最基本的开始:语言选择用python,vex,Houdini作图
<1>
a,求球的体积.半径为4,中心点为0，左断点为-4,右断点为4
import math
radius = 4.000
diameter = radius *2
# sphere r=4 R=8
# this spere is y=sqrt(16-x*x)
# per cylinder volume is PI*r*r*dtx
def sphere_function(xpos,dtx):
return math.sqrt(16.00-xpos*xpos) * math.sqrt(16.00-xpos*xpos)*dtx
# @n is the is this sphere that will split to n piece along xpos
# if sphere r=4, slice is 8, dtx= R/8 = 1
def calculateVolume(n):
v = 0
dtx = diameter/n
for j in range(0,n,1):
xp = j*(diameter/n) - radius #if sphere radius is 4,left plot is -4 ,right plot is 4
v += sphere_function(xp,dtx)
return v
# sphere volume use base function : 3/4 * PI * (R*R*R)
def ruleCacluateVolume(r):
return 4/3.000 * r*r*r
if __name__ == "__main__":
#split a sphere to 20 cylinder
print "use 4 slice :" ,calculateVolume(4)
print "use 8 slice :" ,calculateVolume(8)
print "use 20 slice :" ,calculateVolume(20)
print "use 120 slice :" ,calculateVolume(120)
print "use 200 slice :" ,calculateVolume(200)
python volume.py
use 4 slice : 80.0
use 8 slice : 84.0
use 20 slice : 85.12
use 120 slice : 85.3274074074
use 200 slice : 85.3312
use sphere volume function : 85.3333333333
可以看到和标准体积的球体不差多少。200个切片就很精确了 几乎一样
b,半球体积：
<2>求类似火箭头的曲线体积:
import math
#
# curve function is y=sqrt(x)
# x range->0-5
#
maxRange = 5.0
def clinder_volume(xpos,dtx):
return math.sqrt(xpos*xpos) * dtx
def curve_volume(n):
v = 0.0
dtx = maxRange/n
for x in range(0,n,1):
xpos = x*(maxRange/n)
v += clinder_volume(xpos,dtx)
return v
if __name__ == "__main__":
print curve_volume(15)
<3> 求a和b为什么值，积分的值最大
<4> 梯形法求积分，simpson法求积分
# Trapezoidal
# S = 1/2(y0+ 2y1 + 2y2 + 2y3+...+ 2yn-1 + yn)
def Trapezoidal(down,up,n,func):
if up==down:
return 0.0
h = float(up-down) / float(n)
start = func(down)
end = func(up)
process = 0.0
for dt in xrange(0,n+1,1):
if dt == 0 or dt == n:
continue
process += 2 * func(down + dt * h)
sum = (start + end + process) * (h/2.0)
return sum
# Simpson
# S = h/3(y0 + 4y1 + 2y2 + 4y3 + 2y4 + ... + 2yn-1 + yn)
# func is f(x)
def Simpson(down,up,n,func):
if up==down:
return 0.0
h = float(up-down) / float(n)
start = func(down)
end = func(up)
process = 0.0
for dt in xrange(0,n+1,1):
if dt == 0 or dt == n:
continue
# select the 1 3 5 7 9... index
if dt%2 == 1:
process += 4 * func(down + dt * h)
# select the 2 4 6 8 10... index
if dt%2 == 0:
process += 2 * func(down + dt * h)
sum = (start + end + process) * (h/3.0)
return sum
if __name__ == "__main__":
# part1
# fx = 5x^4 [0,2] n=4
func = lambda x:5*x*x*x*x
T = Simpson(0,2,4,func)
print T
# part2
# fx = x [1,2] n=4
func2 = lambda x:x
T2 = Simpson(1,2,4,func2)
print T2
# part3
# fx = x*x
func3 = lambda x:x*x
T3 = Trapezoidal(1,2,4,func3)
print T3
func4 = lambda x:x*x + 1
T_T4 = Trapezoidal(-1,1,4,func4)
S_T4 = Simpson(-1,1,4,func4)
print T_T4,S_T4
<4> 复习黎曼和 和 定积分关系
5,求椎体体积:
Houdini求出-55
fx = x^2;
则积分为x^3 / 3
上限为y最大值
下限为y最小值
float ptsx[];
float ptsy[];
float ptsz[];
int npt = npoints(0);
resize(ptsx,npt);
resize(ptsy,npt);
resize(ptsz,npt);
for(int i=0;i<npt;i++)
{
vector pos = point(0,"P",i);
ptsx[i] = pos.x;
ptsy[i] = pos.y;
ptsz[i] = pos.z;
}
float maxy = max(ptsy);
float miny = min(ptsy);
printf("%f,%f\n",miny,maxy);
float dttop = pow(maxy,3) / 3.0;
float dtbottom = pow(miny,3) / 3.0;
float volume = dttop - dtbottom;
adddetailattrib(geoself(),"cvolume",volume);
6,一个立方体x=0 和 x=4 出垂直于x轴的两个平面之间，在0<= X <= 4 垂直于x横截面都是正方形，并且他们对角线都是从抛物线y = -sqrt(x) 和 y = sqrt(x)。
如图:
对角线长度则为2sqrt(x)
对角线一半为d = sqrt(x)
要求变长 h , 已知sina = d / h , 因为a = 45,所以 h =( 2sqrt(x) ) / sqrt(2)
A(x) = h^2 = 2x
求积分2x 0 <=x <= 4
F(x) = x^2
F(4) - F(0) = 16
7,x = sqrt(5) y^2 的曲线（0<y<2），从曲线Y到这条曲线形成的立体，由圆盘组成。
求这个形体体积.
8,
区域有y = x ^2 + 1,y = x+3围成的面积 沿着X轴向旋转，求旋转体体积，。
两线交点:-1 , 2
PI * R(x) ^2 - PI * r(x)^2 的积分.
PI(x+3)^2 - PI(x^2+1)^2 = PI[ (x+3)^2 - (x^2+1)^2 ]
-1<x<2
求积分.
9,y=0与y=5 之间的y = x^2 / 2
a,图像绕Y旋转一周所形成的碗状体积。
b,并且求如果每秒3立方单位的常数速率往碗里灌水，当水深为4个单位时，水面上升的速率。
旋转法求体积，因为绕Y旋转，所以半径是x = sqrt(2y)
面积:PI * r^2 = 2 *PI *y
求积分0,5 区间 , 2*pi*y dy 的积分 是25PI
v(h) = | A(h) dh
则dv/dh = A(h) = 2 * PI * y
dv/dt = (dv/dh) * (dh/dt)
dh/dt 则是我们的速率。 dh/dt = (dv/dt) * (1 / 2*PI*y )
则速率：dh/dt = 3 * ( 1 / 2*PI*4 )
5章 5.2 7题
y=x, y=-x/2 , x=2 求两条曲线 和给定的范围 ，沿着Y旋转的体积。（圆柱薄壳法）
圆柱薄壳法:
微分:根据二阶导y''和一阶导y'大概画函数图像。
<1>
x<2 y'<0 ,y'' <0 2 y=1 y'=0 ,y''<0 2<x<4 y'>0 ,y''<0 4 y=4 4<x<6 y'>9,y''<0 6 y=7 x>6 y'<0.y''<0
。。
来源：https://www.cnblogs.com/gearslogy/p/6831501.html