Two Cylinders （辛普森公式处理积分）

迈不过友情╰ 2022-09-22 03:44 141阅读 0赞

Two Cylinders （辛普森公式处理积分）：[http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120113\#problem/I][http_acm.hust.edu.cn_vjudge_contest_view.action_cid_120113_problem_I] 传送门：nefu

题面描述:

I - Two Cylinders

**Time Limit:**1000MS **Memory Limit:**64000KB **64bit IO Format:**%lld & %llu

Submit [Status][]

Description

In this problem your task is very simple.

Consider two infinite cylinders in three-dimensional space, of radii R1 and R2 respectively, located in such a way that their axes intersect and are perpendicular.

Your task is to find the volume of their intersection.

Input

Input file contains two real numbers R 1 and R 2 (1 <= R 1,R 2 <= 100).

Output

Output the volume of the intersection of the cylinders. Your answer must be accurate up to 10 \-4.

Sample Input

1 1

Sample Output

5.3333

题目大意：

此题要计算两个圆柱体垂直相交之后形成的体积，且两个圆柱体的半径不一样。

题目分析：

可以先依据两个半径相等的圆柱体相交所得物体（称为：牟合方盖）的体积计算方法来推导半径不相等的物体的体积，计算牟合方盖的体积时，我们都是利用定积分，先计算在第一卦限中的体积，最后乘以8就行了，具体过程如下。

求解两柱体x2+y2=R2和x2+z2=R2相交部分的体积。

![Center][]![Center 1][]![1410471108297755055.jpg][]

![Center 2][]

![Center 3][]

而此题所要求的是柱体x2+y2=R22和x2+z2=R12相交的体积。

我们同样也是只算第一卦限的体积，最后乘以8即可。

![Center 4][]

代码实现如下：

#include <iostream>
    #include <cstdio>
    #include <cmath>
    
    using namespace std;
    
    const double esp=1e-10;
    double r1,r2;
    double f(double x)
    {
        return sqrt((r1*r1-x*x)*(r2*r2-x*x));
    }
    
    double Simpson(double a,double b)  
    {
        return (b-a)*(f(a)+f(b)+4*f((a+b)/2.0))/6.0;
    }
    
    double Midd(double a,double b) ///迭代求解数值积分
    {
        double mid=(a+b)/2.0;
        double res=Simpson(a,b);
        if(fabs(res-Simpson(a,mid)-Simpson(mid,b))<esp)
        {
            return res;
        }
        else return Midd(a,mid)+Midd(mid,b);
    }
    
    int main()
    {
    
        while(scanf("%lf%lf",&r1,&r2)!=EOF)
        {
            if(r1==r2)
                printf("%.4lf\n",16.0/3*r1*r1*r1);
            else
            {
                if(r1>r2)
                {
                    double t=r1;
                    r1=r2;
                    r2=t;
                }
                printf("%.4lf\n",8.0*Midd(0,r1));
            }
        }
        return 0;
    }

[http_acm.hust.edu.cn_vjudge_contest_view.action_cid_120113_problem_I]: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120113#problem/I
[Status]: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120113#status//I/0
[Center]: https://img-blog.csdn.net/20160626163832130?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center
[Center 1]: https://img-blog.csdn.net/20160626163844992?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center
[1410471108297755055.jpg]: /images/20220721/7aff590711b34c45994d8938d00a01a8.png
[Center 2]: /images/20220721/c8011a782c864d35bc68229efd04ebbe.png
[Center 3]: /images/20220721/25a6b59ac2304d85b3e87492d776f871.png
[Center 4]: /images/20220721/91d04b9db17f4dc3b7652f68468c3e2d.png