# Investigating ratio of areas and volumes

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Introduction

Investigating Ratios of Areas and Volumes

Michael Zuber

Introduction:

This paper has been separated into three parts:

Part i investigates the ratio of the areas formed when y = xnis graphed between two arbitrary parameters x = a and x=b such that a<b.

Part ii investigates the ratio of the volumes formed around the x-axis when y = xn is graphed between two arbitrary parameters x = a and x=b such that a<b.

Part iii investigates the ratio of the volumes formed around the y-axis when y = xn is graphed between two arbitrary parameters x = a and x=b such that a<b.

To investigate these ratios the program Autograph 3.20 was used in order to obtain sets of results. All areas were found using Simpson’s rule at 50 divisions.

Part i:

Introduction:

In this investigation the ratio of the areas formed when y = xnis graphed between two arbitrary parameters x = a and x=b such that a<b will be investigated.

This investigation will investigate the ratio area A: area B.

Area A is the area contained in between the graph and the y-axis between the arbitrary parameters x = a and x = b.

Area B is the area contained in between the graph and the x-axis between the arbitrary parameters x = a and x = b.

However, in investigating this there is a small problem. Using modern technology the area between the graph and the x-axis can easily be found. However, not all programs allow for the area in between the graph and the y-axis to be found.

This problem, however, can be resolved. Area A can be expressed in another form. Considering the inverse function of y = xn, y = x1/n is found. Considering the graph of y = x1/nwe the following is obtained:

Middle

However, to find Volume A using modern technology is tricky.

Volume A can be expressed in another manner. Consider the horizontal line that meets the graph at the point x = b (y = bn):

Volume A can be considered as the volume contained around the x-axis by the graph of this horizontal line (y = bn)between the points x = 0 and x = b subtracted by volume B and the volume contained around the x-axis by the graph of y = an between the points x = 0 ans x = a. This last volume will be considered as volume C.

This method will be used to find the ratio Volume A: Volume B.

Investigation:

y = x2between x = 0and x = 1:

Consider the graph of y = x2 between the points x = 0 and x = 1.

Volume B will equal the volume contained around the x-axis between the graph and the x-axis between the points x = 0 and x = 1.

Volume A will equal the volume contained around the x-axis between the graph of y = 12 (= 1) and the x-axis between the points x = 0 and x = 1 subtracted by volume B. (Note: since in this case a = 0 volume C = 0).

Using the program Autograph 3.20 the following results are obtained:

Volume B = 0.2π

Volume under y = 1 = 1π

Volume A = 1 - 0.2π = 0.8π

Ratio volume B: volume A = 0.8π: 0.2π

= 4: 1

y = xnbetween x = 0and x = 1:

Now consider other functions of the type y = xn, n Z+ between the points x = 0 and x = 1.

Volume B will be the volume contained around the x-axis between the graph and the x-axis between the points x = 0 and x = 1.

Volume A will be the volume contained between the line y = 1n (= 1) and the x-axis between the points x = 0 and x = 1 subtracted by volume B (volume C = 0).

Conclusion

n | Area under horizontal line (in π) | Volume A (in π) | Volume B (in π) | Volume A / Volume B |

2 | 16 | 8.00 | 8.00 | 1.0 |

3 | 32 | 19.20 | 12.80 | 1.5 |

4 | 64 | 42.65 | 21.35 | 2.0 |

5 | 128 | 91.35 | 36.65 | 2.5 |

6 | 256 | 191.70 | 64.30 | 3.0 |

7 | 512 | 397.50 | 114.50 | 3.5 |

8 | 1024 | 817.40 | 206.60 | 4.0 |

9 | 2048 | 1671.00 | 377.00 | 4.4 |

10 | 4096 | 3404.00 | 692.00 | 4.9 |

π | 35.3 | 21.56 | 13.74 | 1.57 |

e | 26.32 | 15.16 | 11.16 | 1.36 |

From these results it is evident that the conjecture holds true between the points x = 0 and x = 2.

y = xnbetween the points x = 1 and x = 2:

Volume A will equal the volume contained around the x-axis between the graph of y = x1/n and the x-axis between the points x = 1n (= 1)and x = 2n.

Volume B will equal the volume contained around the x-axis between the graph of y = 2 and the x-axis between the points x = 0 and x = 2n subtracted by volume A and volume C.

Volume C will equal the volume contained around the x-axis between the graph of y = 1 and the x-axis between the points x = 0 and x = 1n (=1).

n | Area under horizontal line (in π) | Volume A (in π) | Volume B (in π) | Volume C (in π) | Volume A / Volume B |

2 | 16 | 7.5 | 7.5 | 1 | 1.0 |

3 | 32 | 18.6 | 12.4 | 1 | 1.5 |

4 | 64 | 42.66 | 20.34 | 1 | 2.1 |

5 | 128 | 91.41 | 35.59 | 1 | 2.6 |

6 | 256 | 191.9 | 63.1 | 1 | 3.0 |

7 | 512 | 398 | 113 | 1 | 3.5 |

8 | 1024 | 818.6 | 204.4 | 1 | 4.0 |

9 | 2048 | 1674 | 373 | 1 | 4.5 |

10 | 4096 | 3410 | 685 | 1 | 5.0 |

π | 35.3 | 21.56 | 12.74 | 1 | 1.69 |

e | 26.32 | 15.17 | 10.15 | 1 | 1.49 |

From these results it is evident that the conjecture holds true between the points x = 1 and x = 2.

y = xnbetween the points x = e and x = π:

Volume A will equal the volume contained around the x-axis between the graph of y = x1/n and the x-axis between the points x = enand x = πn.

Volume B will equal the volume contained around the x-axis between the graph of y = π and the x-axis between the points x = 0 and x = πn subtracted by volume A and volume C.

Volume C will equal the volume contained around the x-axis between the graph of y = e and the x-axis between the points x = 0 and x = en.

n | Area under horizontal line (in π) | Volume A (in π) | Volume B (in π) | Volume C (in π) | Volume A / Volume B |

2 | 97.41 | 21.41 | 21.40 | 54.6 | 1.0 |

3 | 306 | 94.56 | 63.04 | 148.4 | 1.5 |

4 | 961.3 | 372.00 | 185.93 | 403.4 | 2.0 |

5 | 3020 | 1374.00 | 549.56 | 1097 | 2.5 |

6 | 9488 | 4881.00 | 1626.23 | 2981 | 3.0 |

7 | 29807 | 16880.00 | 4824.84 | 8102 | 3.5 |

8 | 93642 | 57300.00 | 14317.58 | 22024 | 4.0 |

9 | 294185 | 191700.00 | 42616.50 | 59869 | 4.5 |

10 | 924210 | 634600.00 | 126870.13 | 162740 | 5.0 |

π | 359.9 | 115.40 | 73.50 | 171 | 1.57 |

e | 221.7 | 63.19 | 46.51 | 112 | 1.36 |

From these results it is evident that the conjecture holds true between the points x = e and x = π.

Proving the conjecture:

Using calculus the conjecture can be proven.

Volume A:

Volume B:

Ratio volume A: volume B:

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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