• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Investigating Ratios of Areas and Volumes

Extracts from this document...



        In this portfolio, I will investigate the ratio of areas formed when y = xn is graphed between two arbitrary parameters x = a and x = b such that a < b.

Ratio of Area

        Given the function y = x2, I will first consider the region formed by this function from

x = 0 to x = 1 and the x-axis. The region will be labeled B. The region from y = 0 to y = 1 and the y-axis will be labeled A. A diagram of this is show below:


Finding the ratio of area A: area B:

By integrating this function from x = 0 to x = 1, we can find the area under the curve, area B.

B = image05.png = image39.png = image48.png − 0 = image48.png

Since the total area of area A and B is 1, we can simply subtract the area of B from 1 to get area A.

A = 1 − image48.png = image58.png

Thus the ratio of area A: area B is 2:1.

...read more.


1 − image58.png = image48.png

1:2 = ½

y = x5/3

image59.png = image60.png = image61.png − 0 = image61.png

1 − image61.png = image62.png

5:3 = 5/3

The graphs of these functions are shown below:


One important subset of real numbers has been excluded from further consideration is negative numbers. This is because no corresponding y-value exists at x = 0. There is a vertical asymptote at x = 0 and a horizontal asymptote at y = 0, as they approach infinity. Thus y = x-ncannot be integrated as infinity is not a valid area.  The graph below demonstrates them phenomena for

y = x-2:


To further test the conjecture, I will now consider the areas between other x-values. I will examine if the conjecture holds for x = 0 and x = 2, x = 1 and x = 2, and x = 2 and x = 5.

First, I will consider functions of the type y = xn, image54.png between x = 0 and x = 2. The conjecture seems to hold for the bounds x = 0 and x = 2.


Area of B (under the curve)

Area of A


y = x0

image65.png = 2


0:2 = 0

y = x2

image66.png = image67.png = image68.png − 0 = image68.png

8 − image68.png = image69.png


y = x2/3

image70.png= image71.png= 1.9048… − 0 = 1.9048…

image72.png − 1.9038… = 1.2699…

1.2699…/ 1.9038… = 2/3


Area of A and B is now defined as the following:

Area A: y = xn, y = an, y = bn, and the y-axis

Area B: y =x, x = a, y = b, and the x-axis

The graph below is an example of the bounds for y = x2 from x=1 to x=2image74.pngimage02.pngimage03.pngimage00.pngimage01.png

Next, I will consider functions of the type y = xn, image54.png between x = 1 and x =2. The conjecture seems to hold for the bounds x = 1 and x = 2.


Area of B (under the curve)

Area of A


y = x0

image75.png = 1


0:1 = 0

y = x2

image76.png = image77.png = image68.png − image48.png = image78.png

7 − image78.png = image79.png

14:7 = 2

y = x2/3

image80.png=image81.png= 1.9048… − image82.png = 1.3048…

(image72.png−1)  − 1.3048…= 0.8699…

0.869…/ 1.3048… = 2/3

...read more.


y = xn from x = a to x = b such that a < b and for the regions defined below:

Area A: y = xn, y = an, y = bn, and the y-axis

Area B: y =x, x = a, y = b, and the x-axis


B = image92.png = image93.png = image51.pngimage94.png

The formula for area A can be defined as: image95.png

image96.png = image97.png= image98.png=  image99.pngimage07.png= image08.png

Thus the ratio of A: B is n:

image09.png=image10.png= n

Ratios of Volume

After considering and proving that the ratio of the area A: area B is simply n, I will now consider a general formula for the ratios of the volumes of revolution generated by the regions A and B when they are rotated about the x-axis and y-axis.

Proof for x-axis conjecture:

I will use the disc method to solve for volumes of revolution: V = image11.png, y is the function

y = xn

VB = image11.png= image12.png= image13.png= image13.png=image14.png

VA = image15.png= image16.png=image17.png

Thus image18.png=image19.png= image20.png

Proof for the y-axis:

VA = image21.png=image22.png=image23.png=image24.png=image25.png =image26.png

VB = image27.png=image28.png=image29.png= image30.png

image31.png=image32.png =image33.png


Thus the ratio of area A: area B is image20.pngfor the x-axis and image33.pngfor the y-axis.

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    5.94:1 where are close to 6:1 I have found the odds of Adam winning the game and this is a very specific case, in order to generalize the model, the number must be replaced with apt variables. Now I will generalize this model to any player C's probability of winning

  2. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    If we look at the first equation, then the first thing to do in that equation would be to split the fractions so that the equation now looks like: . Once that has been done we can now factorize the brackets' section by , and therefore the equation would look like: .

  1. Investigating the Ratios of Areas and Volumes around a Curve

    A limit clearly cannot be set at x=0, but we must examine whether a pattern holds for these graphs also. As seen in the diagram above, whereas for positive n the areas are separate, for negative n they overlap (the dotted area above).

  2. Investigating ratio of areas and volumes

    Area B is the area subtended in between the graph of y = xn and the x-axis between the two arbitrary points x = a and x = b such that a<b. Investigation: y = x2 between x = 0 and x = 1: Consider the graph y = x2

  1. Shady Areas. In this investigation you will attempt to find a rule to approximate ...

    (3.64 + 4.0) = 0.764 TA= AA + AB + AC + AD + AE TA= 0.604 + 0.62 + 0.652 + 0.7 + 0.764 TA= 3.34 With the help of technology, create diagrams showing and increasing number of trapeziums. For each diagram, find the approximation for the area.

  2. Matrix Binomials ...

    = Using information from above, we can test to see whether or not it is applicable when finding Zn. Essentially, since Xn = 2n-1(X) and Yn = 2n-1(Y), we'll test to see if Zn = 2n-1(Z). When n = 4 Z4 = = Verifying with Zn = 2n-1(Z), Z4 =

  1. Approximations of areas The following graph is a curve, the area of this ...

    = X2 + 3. Here the number of trapeziums are 5 so the M values change, I will show an equation to monitor this data soon. Table 2 M N 0 3 0.2 3.04 0.4 3.16 0.6 3.36 0.8 3.64 1 4 Again I repeat the process of Graph 1 but instead of adding 2 trapezoids, I add 5.

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    of base ×âxn+x value of first base However, the position of the 1st base will be assumed to be in position 0, while the position of the 2nd base is assumed to be in position 1, etc. From the previous graphs, using the obtained values from the table, this seems to be a consistent pattern.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work