Investigating Ratios of Areas and Volumes - Speech or Presentation Example

A graphical depiction of the said areas is shown below: Knowing the value of the area B, the area A may also be computed. Instead of integrating the function, it is simpler to subtract the area below the curve from the total area of the rectangle. This results to area A as shown: By comparing the plots of two curves, it can be seen that the area under the curve decreases as the exponent increases. Consequently, the ratio of the two areas also increases. This trend supports the calculated data.

This shows that for the section of the curve from x = 0 to 1, the ratio is just equal to the exponent. However, does this conjecture hold true for all ranges? To examine this, we reevaluate the generic function y = xn with various ranges: With a range of 0 to 2 for x, the ratio of the areas remains equal to n. However, only the upper limit has been changed in this particular case. By adjusting the lower limit, a certain area is removed from the rectangle as depicted in the following graph. Again, the resultant ratio is simply equal to the exponent n.

In fact, it can be proven that the ratio will remain constant despite any changes in the limit. With a lower limit of a and an upper limit of b, the proof will appear as shown. Now that it is clear that the ratio of the two areas will remain constant, it is possible to extend the analysis to a three-dimensional one by revolving the surface around a specified axis. First, we will investigate the effects of a revolution around the x-axis. Since the area being revolved is that of B, the resulting volume is taken by evaluating the previous expression.

The radius in such a case would be equivalent to y or generally xn. Using this, the volume can be determined in a straightforward manner. The remaining volume is that of the revolution of surface A. However, instead of integrating, it is