def calculate_sphere_volume_m3(radius_m: float) -> float: """ TODO: write the function purpose, parameters, and return value description here """ pi = 3.141592 volume = (4.0 / 3.0) * pi * (radius_m ** 3) return volume def calculate_diff_sphere_volume_m3(radius_m_s1: float, radius_m_s2: float) -> float: """ calculate_diff_spheres_volume_m3(radius_m_s1, radius_m_s2) calculates the difference in volume between two spheres (s1 and s2) using calculate_sphere_volume_m3(). S2 is assumed to be the larger sphere, so this function will basically return volume_m3_s2 - volume_m3_s1. :example: calc_diff_sphere_volume_m3(1.0, 2.0) -> 29.32 (roughly) TODO: add additional examples here :param radius_m_s1: The radius in meters of sphere 1 :param radius_m_s2: The radius in meters of sphere 2 :returns: The difference in volume (cubic meters) of sphere 2 minus sphere 1 """ pass # TODO: replace 'pass' with the correct calls to calculate_sphere_volume_m3()
def calculate_sphere_volume_m3(radius_m: float) -> float: """ TODO: write the function purpose, parameters, and return value description here """ pi = 3.141592 volume = (4.0 / 3.0) * pi * (radius_m ** 3) return volume def calculate_diff_sphere_volume_m3(radius_m_s1: float, radius_m_s2: float) -> float: """ calculate_diff_spheres_volume_m3(radius_m_s1, radius_m_s2) calculates the difference in volume between two spheres (s1 and s2) using calculate_sphere_volume_m3(). S2 is assumed to be the larger sphere, so this function will basically return volume_m3_s2 - volume_m3_s1. :example: calc_diff_sphere_volume_m3(1.0, 2.0) -> 29.32 (roughly) TODO: add additional examples here :param radius_m_s1: The radius in meters of sphere 1 :param radius_m_s2: The radius in meters of sphere 2 :returns: The difference in volume (cubic meters) of sphere 2 minus sphere 1 """ pass # TODO: replace 'pass' with the correct calls to calculate_sphere_volume_m3()
C++ for Engineers and Scientists
4th Edition
ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Bronson, Gary J.
Chapter6: Modularity Using Functions
Section6.2: Returning A Single Value
Problem 13E
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Question
def calculate_sphere_volume_m3(radius_m: float) -> float:
"""
TODO: write the function purpose, parameters, and return value description here
"""
pi = 3.141592
volume = (4.0 / 3.0) * pi * (radius_m ** 3)
return volume
def calculate_diff_sphere_volume_m3(radius_m_s1: float, radius_m_s2: float) -> float:
"""
calculate_diff_spheres_volume_m3(radius_m_s1, radius_m_s2) calculates the difference
in volume between two spheres (s1 and s2) using calculate_sphere_volume_m3().
S2 is assumed to be the larger sphere, so this function will basically return
volume_m3_s2 - volume_m3_s1.
:example: calc_diff_sphere_volume_m3(1.0, 2.0) -> 29.32 (roughly)
TODO: add additional examples here
:param radius_m_s1: The radius in meters of sphere 1
:param radius_m_s2: The radius in meters of sphere 2
:returns: The difference in volume (cubic meters) of sphere 2 minus sphere 1
"""
pass # TODO: replace 'pass' with the correct calls to calculate_sphere_volume_m3()
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