決定把例題用程序都完成一遍。從最基本的開始:語言選擇用python,vex,Houdini作圖
<1>
a,求球的體積.半徑為4,中心點為0，左斷點為-4,右斷點為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
可以看到和標准體積的球體不差多少。200個切片就很精確了 幾乎一樣
b,半球體積：
<2>求類似火箭頭的曲線體積:
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為什么值，積分的值最大
<4> 梯形法求積分，simpson法求積分
# 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> 復習黎曼和 和 定積分關系
5,求椎體體積:
Houdini求出-55
fx = x^2;
則積分為x^3 / 3
上限為y最大值
下限為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);
6,一個立方體x=0 和 x=4 出垂直於x軸的兩個平面之間，在0<= X <= 4 垂直於x橫截面都是正方形，並且他們對角線都是從拋物線y = -sqrt(x) 和 y = sqrt(x)。
如圖:
對角線長度則為2sqrt(x)
對角線一半為d = sqrt(x)
要求變長 h , 已知sina = d / h , 因為a = 45,所以 h =( 2sqrt(x) ) / sqrt(2)
A(x) = h^2 = 2x
求積分2x 0 <=x <= 4
F(x) = x^2
F(4) - F(0) = 16
7,x = sqrt(5) y^2 的曲線（0<y<2），從曲線Y到這條曲線形成的立體，由圓盤組成。
求這個形體體積.
8,
區域有y = x ^2 + 1,y = x+3圍成的面積 沿着X軸向旋轉，求旋轉體體積，。
兩線交點:-1 , 2
PI * R(x) ^2 - PI * r(x)^2 的積分.
PI(x+3)^2 - PI(x^2+1)^2 = PI[ (x+3)^2 - (x^2+1)^2 ]
-1<x<2
求積分.
9,y=0與y=5 之間的y = x^2 / 2
a,圖像繞Y旋轉一周所形成的碗狀體積。
b,並且求如果每秒3立方單位的常數速率往碗里灌水，當水深為4個單位時，水面上升的速率。
旋轉法求體積，因為繞Y旋轉，所以半徑是x = sqrt(2y)
面積:PI * r^2 = 2 *PI *y
求積分0,5 區間 , 2*pi*y dy 的積分 是25PI
v(h) = | A(h) dh
則dv/dh = A(h) = 2 * PI * y
dv/dt = (dv/dh) * (dh/dt)
dh/dt 則是我們的速率。 dh/dt = (dv/dt) * (1 / 2*PI*y )
則速率：dh/dt = 3 * ( 1 / 2*PI*4 )
5章 5.2 7題
y=x, y=-x/2 , x=2 求兩條曲線 和給定的范圍 ，沿着Y旋轉的體積。（圓柱薄殼法）
圓柱薄殼法:
微分:根據二階導y''和一階導y'大概畫函數圖像。
<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
。。