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：

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：

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我们先求旋转体的体积
我们首先可以写出圆的公式

$$(x-r)^2 + y^2 = d^2$$
所以得到 $$x = f(y) = \sqrt {d^2-y^2} + r$$
由公式 $$V = \pi \int_0^d \pi x^2 dy$$ 得到
$$V_1 = \pi \int_0^d(\sqrt{d^2-y^2} + r)^2dy$$
$$= \pi \int_0^d(d^2-y^2+r^2+2r\sqrt{d^2-y^2})dy$$
$$= \pi d^3 + \pi r^2d-\frac{1}{3}\pi d^3+\pi \int_0^d2r\sqrt{d^2-y^2}dy$$
$$= \pi r^2d+\frac{2}{3}\pi d^3+\pi \int_0^d2r\sqrt{d^2-y^2}dy$$
令 $$y = dsin\theta$$
$$\int_0^d \sqrt{d^2-y^2} = \int_0^{\frac{\pi}{2}} dcos\theta ddsin\theta = \int_0^{\frac{\pi}{2}} d^2cos^2\theta d\theta$$
$$= d^2 \int_0^{\frac{\pi}{2}} \frac{cos2\theta + 1}{4} d2\theta$$
$$= d^2 \int_0^{\pi} \frac{cost + 1}{4}dt$$
$$= \frac{1}{4}\pi d^2$$
所以 $$V_1 = \pi r^2d+\frac{2}{3}\pi d^3 + \frac{1}{4}\pi d^2$$
$$V_2 = \pi (r+d)^2h$$
$$V = 2V_1 + V_2$$
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 $$S = 2\pi \int_a^bf(x)\sqrt{1+f'(x)^2} dx$$
所以
$$S_1 = 2\pi \int_0^d(\sqrt{d^2-y^2}+r)\sqrt{1+\frac{y^2}{d^2-y^2}}dy$$
$$= 2\pi d^2+\pi^2 dr$$
上下两个圆
$$S_2 = 2\pi r^2$$
侧面积
$$S_3 = 2\pi(r+d)h$$
$$S = S_1 + S_2 + S_3$$
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;
}