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Investigating Ratios of Areas and Volumes

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Introduction

                Devan

Bob Devan

Calculus AB/BC

7th & 8th

December 6, 2008

Investigating Ratios of Areas and Volumes

1.Given the function y=image35.png, consider the region formed by this function from image69.png=0 to image69.png=1 and the image69.png-axis. Label this area B. Label the region from y=0 to y=1 and the y-axis area A.image34.png

A. Find the ratio of area A: area B.image00.pngimage01.png

The area of A for the given function of image89.pngimage35.pngisimage100.png. image12.png

Area A image106.png

The area of B for the given function of image89.pngimage35.png is image41.png

Area B:image47.png

Thus the ratio of area A : area B can be given as 2:1.

B. Calculate the ratio of the areas for other functions of the type image53.png between image59.png0 and image65.png. Make a conjecture and test your conjecture for other subsets of the real numbers.

Area A: 1-image66.png

Area B:  image67.png

Conjecture: Given the function image53.png

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Middle

 is image88.pngimage88.png

Area A:image90.pngimage90.png

The area of B for the function image91.pngimage91.png is image92.pngimage92.pngimage33.png

Area B: image93.pngimage93.pngimage02.png

image03.png

image04.png

image05.png

Ex 3)image95.pngimage95.pngimage94.png

Area A: image96.pngimage96.png

Area B:image97.pngimage97.pngimage06.png

image07.png

image08.png

image09.png

3. Is your conjecture true for the general case image98.pngimage98.pngfrom image99.pngimage99.png to image101.pngimage101.png such that image102.pngimage102.png and for the regions defined below? If so prove it; if not explain why not.

Area A:image103.pngimage103.pngand the y-axis

Area B: image104.pngimage104.png and the x-axis

image11.pngimage10.png

Area A:image105.pngimage105.png = image107.pngimage107.png       = image108.pngimage108.png =image109.pngimage109.pngimage13.png

Explanation: The equation of image110.pngimage110.png can be stated in terms of y as image111.pngimage111.png or image112.pngimage112.png. The integral for the area of A is put into terms of y and then solved.image14.png

Area B: image113.pngimage113.png              = image114.pngimage114.pngimage16.pngimage15.png

Explanation: Instead of putting the integral in

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Conclusion

image19.png

πimage48.pngimage48.pngimage49.pngimage49.png = image50.pngimage50.png       = image51.pngimage51.pngimage23.png

Explanation: Region B has no hole in the center thus it can be integrated together without separation.

Conjecture: The ratio of volumes of Region A to Region B is 2n:1. This can be proven by taking the volume of Region A and dividing it by the volume of Region B.

image52.pngimage52.png =image54.pngimage54.pngimage55.pngimage55.png

Region A around y-axisimage24.png

image56.pngimage56.pngimage57.pngimage57.png        = image58.pngimage58.pngimage25.png

Explanation: Since region A around the y-axis is a solid figure there is no need to have multiple integrals. So the volume of region A is simply found by using the disk method.

Region B around y-axisimage26.pngimage26.png

image27.png

image60.pngimage60.png = image61.pngimage61.png     =  image62.pngimage62.png

image28.png

Explanation: Since region B contains a portion cut out, to find the area of B we use the shell method.

Conjecture: My conjecture for the volumes of regions when rotated about the y-axis is the ratio of n:2. This can be proven by taking region A and dividing it by region B.

image63.pngimage63.png=  image64.pngimage64.png

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