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Winston Balmaceda

Mrs. Tallman

AP Calculus

22 February, 2016

Solids Of Revolution

Every American student who pays enough attention in class has some ideas

about area and volume. The two concepts are usually introduced in middle school,

accompanied by simple rules and formulas which can be followed easily. The in-depths

ideas of solving area and volume through multiple means is explored in geometry

classes and the subsequent math classes. The area of a shape, being found using 2-

dimensional objects, and the volume being found using 3-dimensional objects. These

simple rules are used time and time again in mathematics, whether with octagon or

dodecahedrons.

Calculus techniques however can create a more in-depth understanding of the

volume and area of objects in the surrounding world. Calculus allows for one to find the

area under the curve of a graph by simply using the equation of that graph, and calculus

allows one to go further by being able to find the volume and surface area of an item

rotated around an axis, or even by using that function as the base of a cross section.

Using calculus a person would be able to find the volume of any item, whether it’s the

donut you eat every day, or a tiny screw, using calculus techniques many things are

possible.

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In order to find the area between a function and an axis, known as the area

under the curve, two conditions must be met. The function needs to be known, with a

useable equation, and the area to be found must be bound.

Figure 1. The Curve of y=x2

Figure 1, shows a curve. The curve is the graph of y=x2, the area under the

curve is bound from x = 0 to x = 1.5. There are many methods to find the area under the

curve, from counting the 1x1 squares within the boundaries, to using actual calculus. In

this case calculus will be used in order to give a more accurate and faster estimate of

the area under the curve.

The area under a curve can be found using a definite integral,

Area=∫a

b

f ( x )dx

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The integral would give the area under the function f(x) from x = a to x = b. The concept

behind this is simple, the area under the curve is found by finding the product of the

curves x and y values, by cutting the graph into infinitely skinny rectangles, and

summing up the areas of all the rectangles within the bounds.

Figure 2. The Curve y=x2 and the Cuts

Figure 2 above shows the same graph as figure 1, however now included are two

vertical lines, and two horizontal lines. The vertical lines are at x = a and x = b, the

distance between these two lines is the change in x, or dx. If the top of the area found

with dx was horizontal, then the area of the region would be the product of the function

evaluated at a or b and dx, however since the graph given is a curve, there needs to be

another factor. With dx is dy, or the change in y, this is shown in the figure bound by

perpendicular functions to x = a and x = b.

To get the total area underneath the curve for the function shown above from

x = 0 to x = 1.5, every area of the rectangle needs to be added up. However this is not

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simple to perform by hand, so calculus allows for the use and written notation of an

integral as shown:

Area=∫0

1.5

x2dx

This integral would find the area under the curve of f(x) from x = 0 to x = 1.5, to do so

one must find the anti-derivative of the integrand.

Area=∫0

1.5

x2dx

Area=13x3¿0

1.5

Once the integrals anti-derivative is found, the area can be found by subtracting the

greatest bound of the anti-derivative by the lowest bound of the anti-derivative.

Area=131.53−1

303=1.125

Figure 3. Area under f ( x )=x2 from x = 0 to x = 1.5

Figure 3 shows the exact area under the curve f ( x )=x2 from x=0 to x=1.5, which

is 1.1250 units2, rounded to three decimal places.

Using very similar principals, the area between to curves can be found as quickly

as the area under the curve can be found.

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Figure 4. Area Bounded by the Graphs y=x3 and y=√x

In this instance, as shown in figure 4, the area bounded by the curves, notated as

R, is to be calculated. The solid curve is the graph of y=x3 and the dashed line is the

graph of y=√x. The same concept for finding the area under the curve will apply to

finding R. The area bounded will be cut up into small rectangles, similar to how the area

under the curve is cut up. The main difference between these methods is that the height

of the rectangles would be found using the difference in y-values of the two given

functions, rather than of just the y-value of the function. Since the height of the triangles

is the difference of the two functions, the height will be the area just within the bounded

region, with each rectangle having a width of dx, which is still the same change in x.

In order to set up the integral, the same pieces are required. The integral

requires the two functions as well as the bounds they are within. Since the area is being

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found bounded by the two functions, the points of intersection of the two functions are

the bounds. In this case the bounds are at x = 0 and x = 9.

Solving the integral and integrand for the anti-derivative is done in a different

manner however. Because the area bounded by the curves is essentially the difference

between the areas under the two curves, the integral can be set up as such:

Area=∫0

9

√x dx−∫0

9 x3dx

The curve being subtracted from is the curve higher in the graph, in order to give a

difference of areas. Using this the anti-derivative of both integrands can be easily found

and computed. However in the case in which the problem is done using a computation

device of any sort, the integral can be written as a single integral, with the integrand

being the difference in functions as such:

Area=∫0

9

(√ x¿−x3 )dx ¿

However, while it is possible to find the anti-derivative of this integrand, it’s simple to

use the first integral.

Area=∫0

9

√x dx−∫0

9 x3dx

Area=23x3 /2∨¿0

9−16x2∨¿0

9 ¿¿

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Area=23 93 /2

❑

−16 9

2

❑

❑

=4.5units2

Figure 6. Area between y=√x and y=x3

Figure 6 shows the solution to the integral, showing that the area bounded by the

curves, R, is about 4.5 units2.

The definite integral has many uses in area of under the curves of functions, and

it can also be used to find the volume of a function. In order to find the volume using

definite integrals however, the curve must be rotated around an axis. There are three

different ways in which volume can be found using integrals, by shells, disks, and rings.

Each method uses the same concept of finding the volume of various types of cross

sections and adding them in order to find a total volume.

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Figure 7. The Graph of y=x2 Revolved Around the X-Axis with Individual Slice

Figure 7 shows the first method, which is volume by disks. In the disk method a

single graph is rotated around an axis, the axis which it is rotated around determines the

cut taken from the function. The cut for this method is always taken perpendicular to the

axis of revolution. When a dx rectangular cut is revolved around an axis, a cylinder is

created. The volume of all of these 3-Dimensional cylinders is found by the products of

the area of the base and the height of the cylinder. The area of the base is A=π r2, for

dx cuts the r is f(x), and the height of the cut is dx. The integral formula will create an

infinite amount of cylinders, and each volume will be added to the last. The standard

formula for the integral of a disk around the x-axis would be:

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Volume=∫a

b

π (f ( x ))2dx

And for the integral of a disk around the y-axis would be:

Volume=∫a

b

π ( x )2dx

In this example the volume of the solid would be bound from x = 0 to x = 1.5, and

would be shown and integrated as such:Volume=∫0

1.5

π ( x)4dx

Volume=π∫0

1.5

( x )4dx

Volume=π 15x5∨¿0

1.5¿

Volume=π 15 1.55❑

=1.5188π unit s3

Figure 8. Volume by disks

The coefficient π is able to remain outside of the integral because it will be later factored

back into the solution, and the final area is found to be 1.5188 units3 as shown in figure

8.

In the cases in which the axis of rotation is a different axis than the equation, the

way in which the problem is solved changes completely. In order to use volume by

disks, the cut must be perpendicular to the axis, so the integral must be in terms of y.

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Figure 9. The Graph of x=√ y Revolved Around the Y-Axis with Individual Slice

As shown above in figure 9, the cut is dy, and the cylinders have a radius of

x=√ y, which is y=x2 in terms of y. The area of the circular base is the same as the

previous problem, and the integral is set up in the exact same format, but in terms of y.

With that the integral would be as follows:

Volume=π∫0

9

y dy

The integrand is y due to the fact that the radius is squared, and the radius is the square

root of y. The volume is then found through the following:

Volume=π∫0

9

ydy=40.5 πun its3

Figure 10. Volume by Disks around the Y-Axis

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Figure 10 shows the solution of the integral, where the volume of the solid

between the areas of y = 0 to y = 9, and revolving it around the y-axis would result in a

volume of 40.5π units3.

The second method, known as volume by shells, can be used to rotate the curve

y=x2 without changing the terms of the equation. The shells are cylindrical, with a

thickness of dx, can be used to find the volume. The volume of the shell can be found

by rolling each shell out into a rectangular prism.

Figure 11. Shell Method

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When the shell is rolled out, you can see the three dimensions of the shell: the

circumference of the base (2πr), the height (f(x)), and the thickness (dx). The cuts in the

shells method are parallel to the axis of rotation, allowing for the integral to remain in

terms of x. The general equation for shells is as follows:

Volume=2π∫a

b

radius∗height∗thickness

The radius of the base is x, and the height of the shell is the function itself, the

thickness is the cut, in this case dx. The volume of each infinitely hollow shell is used to

sum the total volume. When all of the variables are substituted into the formula, the

volume can be found.

Volume=2π∫0

3

( x ) (x2 )dx=40.5π units3

Figure 12. Volume of Shells Revolved Around the Y-Axis

Figure 12 substituted each variable into its appropriate place, and the volume of

the solid generated is found to be 40.5π units3. The area bounded is different as the

volume is meant to equal the previous problem, since the area is bounded by the x-axis

rather than the y-axis, the bounds are different. The primary use of this method is to

resolve a problem without having to set an equation in terms of y, or if it is not possible

through simple means to put the equation in terms of y.

The final method which can be used to find the volume of a solid is the rings

method. The rings method is used primarily to find the volume of solids that were

formed by taking areas bounded by curves and rotating them about an axis. The axis in

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particular does not need to be the x-axis or y-axis, but rather any horizontal or vertical

axis on a graph. For example we can take the two graphs from the area bounded by the

curves example, y=√x and y=x3 . The area bounded by the graphs is then revolved

around an axis, in this case the y= -2 line will serve as the axis of rotation.

Figure 13. The Region of R reflected over the Axis of Rotation

Figure 13 shows the region R from the areas bounded example reflected over

the axis of rotation at y = -2. Along with this revolution, the rings which will be used to

calculate the volume is formed, each ring cut out is bounded by the area revolved. The

cut for the rings are perpendicular to the axis of rotation. The top function subtracted by

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the lower function would allow for the region between the functions to be calculated, and

the general solution for the problem would be as follows:Volume=π∫a

b

(R2−r2 )dx

In the case of the functions bounded being rotated around the y = -2 axis must be

corrected in order for the integral to correctly calculate the volume of the solid. To do so,

the radii must have an addition such that the large radius is R=√x+2 and the smaller

radius is r=x3+2 to create an integral, the limits of integration match the previous area

bounded by curves problem. The integral is as follows:

Volume=π∫0

9

¿¿

The integral can be simplified through expansion of the binomials and then the anti-

derivative could be found, this method is not required however.

Volume=π∫0

9

¿¿

Figure 14. Volume by Rings

Figure 14 shows that the volume of the solid formed by taking the area between

the curves y=√x and y=x3 and revolving it around the horizontal axis y=−2 is 31.5π

units3.

There is one final method to finding the volume of a solid, known as the cross

section method. With this method, a base is formed by using geometric shapes, or the

area bounded by a function. In order to form a solid, cross sections such as square,

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semi-circles or triangles are taken. Similar to the previous methods, an infinite amount

of cross sections will be summed in order to find the total volume of the solid formed. In

the example to be used, the cross sections are isosceles right triangles perpendicular to

the x-axis, making them dx cuts.

Figure 15. The region R with isosceles right triangle cross-sections

Figure 15 shows the region R with the right triangle cross-sections, the axis are

rotated to give a better view of the triangles. The triangles are infinitely thin prisms with

a thickness of dx. The volume of the prism would be the area of the “base” which is the

right triangle, and the height, which is the thickness, dx. The area of a triangle is

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A=12bh, and since the triangles are right isosceles, the base and height are the same,

meaning the area can simply be described as A=12b2. In this b is this difference of the

two functions values, with this the integral can be written as:

Volume=12∫0

9

¿¿

The bounds are the same for the integral as both of the examples where the area is

bounded. Volume=12∫0

9

¿¿

Figure 16. Volume of Cross-Sections

Figure 16 shows that the volume of the solid when the region between the curves

y=√x and y=x3 is used as the base for a solid with cross-sections of isosceles right

triangles is 1.35 units3.

Calculus allows for many different interesting applications, and with it is the ability

to find the volume of almost any object. No matter the method however, the concepts

are very similar, with the same basic principle of the definite integral being used. Each

concept uses infinitely small slices in order to find the total area or volume, and each

method can be used on a range of items from your morning donut to a block of delicious

cheese.

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