Cylinder Candy ZOJ - 3866 (求定积分)

Cylinder Candy
ZOJ - 3866

 Edward the confectioner is making a new batch of chocolate covered candy. Each candy center is shaped as a cylinder with radius r mm and height h mm.

The candy center needs to be covered with a uniform coat of chocolate. The uniform coat of chocolate is d mm thick.

You are asked to calcualte the volume and the surface of the chocolate covered candy.

Input

There are multiple test cases. The first line of input contains an integer T(1≤ T≤ 1000) indicating the number of test cases. For each test case:

There are three integers r, h, d in one line. (1≤ r, h, d ≤ 100)

Output

For each case, print the volume and surface area of the candy in one line. The relative error should be less than 10-8.

Sample Input

2
1 1 1
1 3 5

Sample Output

32.907950527415 51.155135338077
1141.046818749128 532.235830206285

题意:抽象出来就是求一个圆柱体外面加一层厚度为d的巧克力,求包装完这个立体图形的体积和表面积。

剖析:

以下是包装后的三视图:

这里写图片描述

这里写图片描述

这里写图片描述
所以这道题目的重点就是要求出旋转体的体积和表面积,那么很明显需要用定积分来求,求出后,中间部分就是圆柱形的体积和面积直接可以用公式求得,上下两部分需要用定积分,其推导过程如下:
首先我们需要求旋转体体积和表面积的积分公式如下图:
这里写图片描述
这里写图片描述

这里写图片描述
我们先求旋转体的体积
我们首先可以写出圆的公式

(xr)2+y2=d2

所以得到
x=f(y)=d2y2+r

由公式
V=πd0πx2dy
得到
V1=πd0(d2y2+r)2dy

=πd0(d2y2+r2+2rd2y2)dy

=πd3+πr2d13πd3+πd02rd2y2dy

=πr2d+23πd3+πd02rd2y2dy

y=dsinθ

d0d2y2=π20dcosθddsinθ=π20d2cos2θdθ

=d2π20cos2θ+14d2θ

=d2π0cost+14dt

=14πd2

所以
V1=πr2d+23πd3+14πd2

V2=π(r+d)2h

V=2V1+V2

下面我们来求旋转体的表面积
同样我们由公式如果曲线方程为
y=f(x)

绕x轴旋转一周的表面积
S=2πbaf(x)1+f(x)2dx

所以
S1=2πd0(d2y2+r)1+y2d2y2dy

=2πd2+π2dr

上下两个圆
S2=2πr2

侧面积
S3=2π(r+d)h

S=S1+S2+S3

然后公式推到出来了,直接照着公式写代码就行了
code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const double pi = acos(-1.0);
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        double r,d,h;
        scanf("%lf%lf%lf",&r,&h,&d);
        printf("%.10f ",2.0*(2.0/3.0*pi*d*d*d+pi*r*r*d+1.0/2.0*pi*pi*d*d*r)+pi*(r+d)*(r+d)*h);
        printf("%.10f\n",2.0*(2.0*pi*d*d+pi*pi*d*r)+2.0*pi*r*r+2.0*pi*(r+d)*h);
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/codeswarrior/article/details/80083726