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.
=
