Juice

Once upon a time, there lived a mad programmer. He loved to solve creative problems other than anything. His wife loved him quite a lot but disliked his curiosity for the problems. One day he came from office, his wife gave him a glass of cold lime juice. She was in a romantic mood and waiting for some romantic stuff. But the programmer asked her curiously, "If I give u radius of the top and bottom part of the glass and the height, can you come up with the volume of the glass?" His wife became a bit disappointed but as she is smart she replied with a smile, "You already have drunk some juice, and the glass is not full. If I give you the height of the juice, can you find the volume of the remaining juice in the glass?" Then the programmer kissed his wife and said, "You are the best problem setter in the world!"

Now he set the same problem for you. The radius of the upper part r₁ and lower part r₂ is given. If height of the glass is h and height of the juice is p what is the volume of the juice in the glass?

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing four integers r₁ r₂ h p (1 ≤ r₂ < r₁ ≤ 100, 1 ≤ p ≤ h ≤ 100).

Output

For each case, print the case number and the volume of the juice in the glass. Errors less than 10^-6 will be ignored.

Sample

Sample Input	Sample Output
Click to copy 2 5 2 3 2 5 2 3 3	Click to copy Case 1: 58.643062867 Case 2: 122.52211349

Total Submissions

Users Tried

Users Solved

LOJ 1216 - Juice in the Glass

The radius of the upper part r1 and lower part r2 is given. If height of the glass is h and height of the juice is p what is the volume of the juice in the glass?

Solution

To solve it, me must know how to calculate the volume of a Conical Frustum.

The volume of a conical frustum, V = (π/3) * height * (radius1² + radius2² + (radius1 * radius2)). So, if we know the 2 values of the radii, we can calculate volume of the juice. The only thing we need to figure out is, what is r3, the radius of glass at point p. Since, already r2 and r1 is given, we can find out the value of r3 by doing some tri-gonometric calculations.

In the figure:

In △EGC, 
    tan(α) = EG/EC
           = (r1-r2)/h ...(i)

In △MNC,
    tan(α) = MN/MC
           = (r3-r2)/p
    => r3 - r2 = p*tan(α)
    => r3 = p*tan(α) + r2 
          = p*(r1-r2)/h + r2 [From, (i)]

Now we can just simply plugin the value of r3 in our formula, V = (π/3) * p * (r3² + r2² + (r3 * r2)).

The above solution is accepted.

Solution in C

#include <stdio.h>
#include <math.h>

int main()
{
    double pi, r3;
    pi = 2.0*acos(0.0);
    int t, r1, r2, h, p;
    scanf("%d", &t);
    for (int i = 1; i <= t; i++)
    {
        scanf("%d %d %d %d", &r1, &r2, &h, &p);
        r3 = p/(h * 1.0) * (r1 - r2)  + r2;
        printf("Case %d: %.10lf\n", i, (pi / 3.0) * p * (r3 * r3 + r2 * r2 + (r3 * r2)));
    }
    return 0;
}

Tutorial source