Note: Click on "Kernel" > "Restart Kernel and Run All" in JupyterLab after finishing the exercises to ensure that your solution runs top to bottom without any errors. If you cannot run this file on your machine, you may want to open it in the cloud .

Chapter 2: Functions & Modularization (Coding Exercises)¶

The exercises below assume that you have read the first part of Chapter 2.

The ...'s in the code cells indicate where you need to fill in code snippets. The number of ...'s within a code cell give you a rough idea of how many lines of code are needed to solve the task. You should not need to create any additional code cells for your final solution. However, you may want to use temporary code cells to try out some ideas.

Volume of a Sphere¶

Q1: The volume of a sphere is defined as $\frac{4}{3} * \pi * r^3$. Calculate this value for $r=10.0$ and round it to 10 digits after the comma.

Hints:

use an appropriate approximation for $\pi$
you may use the standard library to do so if you have already looked at the second part of Chapter 2.

In [1]:

import math

In [2]:

r = 10.0

In [3]:

round((4 / 3) * math.pi * r ** 3, 10)

Out[3]:

4188.7902047864

Q2: Encapsulate the logic into a function sphere_volume() that takes one positional argument radius and one keyword-only argument digits defaulting to 5. The volume should be returned as a float object under all circumstances.

In [4]:

def sphere_volume(radius, *, digits=5):
    """Calculate the volume of a sphere.

    Args:
        radius (float): radius of the sphere
        digits (optional, int): number of digits
            for rounding the resulting volume

    Returns:
        volume (float)
    """
    return round((4 / 3) * math.pi * radius ** 3, digits)

Q3: Evaluate the function with radius = 100.0 and 1, 5, 10, 15, and 20 digits respectively.

In [5]:

radius = 100.0

In [6]:

sphere_volume(radius, digits=1)

Out[6]:

4188790.2

In [7]:

sphere_volume(radius)  # or sphere_volume(radius, digits=5)

Out[7]:

4188790.20479

In [8]:

sphere_volume(radius, digits=10)

Out[8]:

4188790.2047863905

In [9]:

sphere_volume(radius, digits=15)

Out[9]:

4188790.2047863905

In [10]:

sphere_volume(radius, digits=20)

Out[10]:

4188790.2047863905

Q4: What observation do you make?

< your answer >

Q5: Using the range() built-in, write a for-loop and calculate the volume of a sphere with radius = 42.0 for all digits from 1 through 20. Print out each volume on a separate line.

Note: This is the first task where you need to use the built-in print() function.

In [11]:

radius = 42.0

In [12]:

for digits in range(1, 21):
    print(digits, sphere_volume(radius, digits=digits))

1 310339.1
2 310339.09
3 310339.089
4 310339.0887
5 310339.08869
6 310339.088692
7 310339.0886922
8 310339.08869221
9 310339.088692214
10 310339.0886922141
11 310339.0886922141
12 310339.0886922141
13 310339.0886922141
14 310339.0886922141
15 310339.0886922141
16 310339.0886922141
17 310339.0886922141
18 310339.0886922141
19 310339.0886922141
20 310339.0886922141

Q6: What lesson do you learn about the float type?

< your answer >