The area of A for the given function of is.

Area A

The area of B for the given function of is

Area B:

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 between 0 and . Make a conjecture and test your conjecture for other subsets of the real numbers.

Area A: 1-

Area B:

Conjecture: Given the function the area of A in ratio of the area of B can be given as n:1.

2. Does your conjecture hold only for areas between x=0 and x=1?

Examine for =0 and =2; =1 and =2 etc.

Ex 1) The area of A for the function y=x2 [0,2] is 8/3

Area A:

The area of B for the function y=x2 [0,2] is 16/3

Area B:

Ex 2) The area of for the function is

Area A:

The area of B for the function is

Area B:

Ex 3)

Area A:

Area B:

3. Is your conjecture true for the general case from to such that and for the regions defined below? If so prove it; if not explain why not.

Area A: and the y-axis

Area B: and the x-axis

Area A: = = =

Explanation: The equation of can be stated in terms of y as or . The integral for the area of A is put into terms of y and then solved.

Area B: =

Explanation: Instead of putting the integral in terms of y, we put the integral in terms of x and then we solve.

Since my conjecture was n:1 which is the same as . So we can state the ratio as

which would be . This would simplify to which would support my conjecture.

4. Are there general formulae for the ratios of the volumes of revolution generated by the regions A and B when they are each rotated about:

(a) x-axis

(b) y-axis

State and prove your conjecture.

Region A around x-axis

Part A. = x = π(

Part B. =

Volume of Region A: π =

π=

Explanation: Region A is simply divided into 2 segments. One being from [0,a] and the other from [a,b]. They are both integrated separately and then added together. After some simplification we get the volume of region A.

Region B around x-axis

π = =

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.

=

Region A around y-axis

=

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-axis

= =

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.

=