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Sydni Jordan
Mr. Acre
AP Calculus
23 February 2015
Solids of Integration
Integration is just one of the few central topics that make up the branch of
mathematics we call Calculus. It is a very broad term considering that there are
numerous applications of topic – such as reverse differentiation and summation.
Although integration is used in many contexts and math problems, this discussion will
focus primarily on the use of integration in the calculations of area and volume of
graphed regions and solids.
Definite integration is a straightforward method that is used to calculate the area
under a curve on an interval. To begin, it must be understood that when the equation of
a function is graphed, there exists a space between the x-axis and said function, f(x). It
is the area of a portion of this region that the definite integral calculates. The shaded
area in Figure 1 shows this region under the curve of a sample equation, 𝑦 = √𝑥.
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Figure 1. The area under the curve of the sample graph.
When calculating the integral of a function, an interval is often included to place
upper and lower boundaries on the designated portion of the area. The x-values that
serve as this boundary are notated as b and a respectively. In the sample graph from
Figure 1, these lower and upper boundaries are set at approximately 0 and 4, indicating
that an integral would evaluate all of the x values within this closed interval of the
sample function.
The calculus notation of the definite integral function is as follows:
𝑏
∫ 𝑓(𝑥) 𝑑𝑥
𝑎
When the equation of the function and its lower and upper x-values are inserted into the
placeholders of the integral function, the area that is bound by the function and the xaxis on that designated interval is computed. However, there is more to it than simply
plugging an equation into a formula and coming up with a numerical value. The symbol,
dx, emphasizes the change in x along the designated interval. To better understand,
imagine slicing the graph into vertical strips that are equal in width. The symbol, dx,
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represents the width of the slices and, therefore, signifies the change in the x values
along the x-axis (see Figure 2).
Figure 2. The changes in x labeled on the sample graph
Because the definite integral function includes the dx and is evaluated on an interval, it
is the sum of all of the areas between a and b. Furthermore, the notation for the definite
integral of the sample equation is as follows:
4
∫ √𝑥 𝑑𝑥
0
Definite integrals calculate not only the area under one curve, but also the area
between two curves. The method used to solve for this area is very similar to the first
approach. To begin, there needs to be a pair of graphed functions. It is essential that
the two equations are graphed because the upper and lower boundaries for the definite
integral are defaulted to where the two graphs intersect each other. For the sample
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𝑥
equations, 𝑦 = √𝑥 and 𝑦 = 3, the intersection points are located at (0, 0) and (9, 3).
See Figure 3 for the graph of these two functions.
Figure 3. The area between two curves.
To find the area between these specific functions, the lower and upper boundaries for
the x values will be set to 0 and 9, respectively, because the shaded area in Figure 3
lies within the closed interval between those points.
To solve for the area between any two curves, the equation of the lower lying
function must be subtracted from that of the upper function. This is the equivalent of
subtracting the region under the lower curve from the total region to solve for the area of
the space that lies between the two. In other words, the equation 𝑦 =
𝑥
3
must be
subtracted from the equation 𝑦 = √𝑥 to create one function to insert into the f(x) portion
of the definite integral. Furthermore, the notation and calculation of the definite integral
is as follows:
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9
∫ √𝑥 −
0
𝑥
𝑑𝑥
3
=
9
= 4.5
2
When the placeholders of the integral are replaced with the appropriate values and
functions, the area of the region between the two curves is computed as 4.5 units 2.
The power of definite integrals transcends the simple ability to calculate areas
under and between graphs. With integrals it is possible to rotate the regions under
curves and not only make transform into 3 dimensional shapes, but also find the
volumes of said shapes. Three methods – disk, ring, and shell – are commonly applied
when performing such a task.
The first method – the disk method – is typically a mode of volume calculation for
solids with dx or dy cuts that are perpendicular to their axes of rotation. In other words,
the disk method is customarily used when the solid that the equation creates either
revolves around the x-axis and uses the dx symbol to signify a change in x values or
revolves around the y-axis and uses the dy symbol mark changes in y values. See
Figure 4 for an example.
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Figure 4. Graph of perpendicular dx cuts and axis
Figure 4 shows that the lines designating the dx cuts on the graph of the sample
equation, 𝑦 = √𝑥, are perpendicular to the x-axis, the axis of rotation. This is a situation
in which the disk method would be used to find the volume of solid of revolution.
In the disk method, each sectioned off slab created by the dx or dy cut – in this
case, dx – forms a circular disk. By finding the volume of each of the disks and adding
them, the definite integral calculates the volume of the entire solid. Figure 5 models a
singular disk in the solid of the rotated sample equation.
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Figure 5. A disk in the solid of revolution.
Because the disk is circular, the equation to find its volume is equivalent to that of a
cylinder, 𝜋𝑟 2 ℎ, where r is the radius of the disk and h is the width. To calculate the
volume of a single disk, the r value must be replaced with the equation of the function
that creates the solid and the designated cut for the solid must take the place of the h
symbol. For the sample, the equation is 𝜋(√𝑥)2 𝑑𝑥. Because the integral is the sum of all
of these disk volumes on an interval, its equation – and the integral of the sample- is as
follows:
𝑏
∫ 𝜋𝑟 2 ℎ
𝑎
𝑏
𝑜𝑟
∫ 𝜋(√𝑥)2 𝑑𝑥
𝑎
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With definite integration through the disk method, the volume of the solid of revolution
may be computed within a designated interval – hence the upper and lower boundaries
indicated by the a and b.
The second method – the ring method – is much like the disk, however it is
typically applied in situations where the area between two curves is rotated around an
axis to form a solid. In the ring method, there are two equations which causes the disk,
once sliced, to have a hole in the center, causing the shape to resemble a ring. See
Figure 6 for a depiction of the ring that is created.
Figure 6. A ring in the solid of revolution.
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Because the ring is also circular, the equation to find its volume is much like that
of the cylinder in the disk method, however the volume of the inner circle that is created
must be subtracted to compute an accurate value. Therefore the equation for the
volume of a singular ring is 𝜋𝑅 2 ℎ − 𝜋𝑟 2 ℎ, where R represents the radius of the outer
circle and r stands for the radius of the inner circle. When applying this equation to a
definite integral, the equation of the upper lying equation must be substituted for the
symbol R while that of the lower equation must be substituted for the r. The upper and
lower boundaries for the interval should be where the two graphs intersect as well.
Much like in the disk method, the h ought to be replaced by the dx or dy cut that is being
used to slice the solid. The equation for the integral that computes the volume is as
follows:
𝑏
∫ 𝜋𝑅 2 ℎ − 𝜋𝑟 2 ℎ
𝑎
However, it is also feasible to rotate such a region about an axis that is not the x(or y-) axis. To complete this task, a minor adjustment must be made to the equation of
the definite integral. When the axis is below the rotating region, the difference between
that axis and the x- (or y-) axis must be added onto the radius of the circles. For
𝑥
example, if the region bounded by the sample equations, 𝑦 = √𝑥 and 𝑦 = 3, is rotated
about the horizontal line 𝑦 = −2 instead of the x-axis, 2 units must be added onto the
radii of the inner and outer circles of the ring. The equation for the definite integral would
then be as follows:
9
2
𝑥
∫ 𝜋(√𝑥 + 2)2 𝑑𝑥 − 𝜋 ( + 2) 𝑑𝑥
3
0
= 31.5𝜋 ≈ 98.96
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Again, the upper and lower boundaries are set to 0 and 9 because these are the x
values that correlate with the points in which the two graphs intersect. The volume of
the solid of revolution created by the rotated region between the two curves is computed
as approximately 98.96 units3.
The third method – the shell method – does not quite follow the patterns of disk
and ring. It is typically a mode of volume calculation for solids with dx or dy cuts that are
parallel to their axes of rotation. In other words, the shell method is customarily used
when the solid that the equation creates either revolves around the x-axis and uses the
dy symbol to signify a change in x values or revolves around the y-axis and uses the dx
symbol mark changes in y values. The sections in the shell method are best understood
if thought of as a tube or hollow cylinder that is cut along the side to lay flat. See Figure
7 for an illustration of a shell section.
Figure 7. A shell in the solid of revolution.
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The equation to find the volume of a singular shell like the one shown in Figure 7
is 2𝜋(𝑟)(ℎ𝑒𝑖𝑔ℎ𝑡)(𝑤𝑖𝑑𝑡ℎ), where the r represents the radius of the shell. To compute the
volume, the equations of the functions should be inserted into the equation for the
height of the shell and the corresponding dx or dy should replace the width. Like the
previous two methods, the equation of the lower function should be subtracted from that
of the upper as it is incorporated into the volume equation as the height of the shell.
Also, the variable r must be replaced with whatever variable or function that dictates the
width of the shell. In the case of the two sample equations, this variable is simply x. The
equation of the definite integral that evaluates the total volume of this solid of revolution
created by the two sample functions is as follows:
9
𝑥
∫ 2𝜋(𝑥) (√𝑥 − ) 𝑑𝑥
3
0
The disk, ring, and shell methods are but three processes in which the volume of
a solid of revolution may be computed. However, there are times when one is required
to find the volume of a solid that is not formed by the rotation of a function or two about
a line. Such figures take the shapes of the most common of geometrical shapes, such
as cubes, wedges, spheres, etc. There is one last technique that has the ability and
versatility to calculate the volume of any designated solid. Said technique is called the
cross-section method.
In this method, the shape of the cross section is dictated by the overall shape of
the solid. Moreover, the computed volume may vary because the cross-sectional shape
is determined by the mathematician.
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Similar to the previous processes, the region under or between two curves may
be incorporated into the volume of the solid. In many cases, said region represents one
face of the solid. For instance, the area between the two sample equations, 𝑦 = √𝑥 and
𝑥
𝑦 = 3, will serve as the base of a solid made with isosceles right triangles. See Figure 8
for a diagram.
Figure 8. Slabs in the cross-section method.
In Figure 8, the base legs of the isosceles triangles span the region bounded by the
sample functions. The complete solid is composed of multiple isosceles triangles such
as those depicted in the figure above. Each triangle has varying lengths for their legs
because the distance between the two functions vary along the interval. The integral,
like in the previous methods of volume calculation, is the sum of all of these crosssectional triangles. Therefore, to find the volume of the complete solid formed by the
equations, the volumes of each and every triangle along this interval must be
compounded into one value.
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To calculate this value for any singular cross-section, the formula for the volume
of the shape of that cross section should be considered. For example, the volume
equation for a triangle would be best to use in the sample problem. This equation is as
follows:
1
𝑉𝑜𝑙𝑢𝑚𝑒 = (𝑏𝑎𝑠𝑒)(ℎ𝑒𝑔ℎ𝑡)(𝑤𝑖𝑑𝑡ℎ)
2
Because the slabs used for the cross-sections are isosceles triangles, the lengths for
both legs are congruent and, therefore, simply squaring the value of the base will be a
more straightforward approach.
The input value for the base of the triangle correlates to the region of the graph
𝑥
that lies between the two curves, 𝑦 = √𝑥 and 𝑦 = 3. However, because the length of
this area varies the equations of the functions will be substituted into the volume
function instead. The dx or dy cut should fill the placeholder for width as well. Much like
the integrals of the past methods of volume calculation, the lower and upper boundaries
for the interval are set to 0 and 9 respectively. The equation and solution for this integral
is as follows:
9
∫
0
1
𝑥
(√𝑥 − )2 𝑑𝑥
2
3
= 1.35
The value correlating to the volume of the entire solid formed by the region between the
two sample curves is computed as approximately 1.35 units3.
To conclude, there are multiple methods and approaches to use when calculating
the area or volume of areas under and between curves. From disk to ring to shell to
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cross-section, the only necessities for accurate computations of these values are the
equations themselves and the knowledge behind the application. None of the
aforementioned methods, however, would have been useful without definite integration,
and yet these are still but a few applications of the function.
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Works Cited
"Definite Integral." Wolfram Alpha: Computational Knowledge Engine. WolframAlpha,
n.d. Web. 20 Feb. 2015.
<http://www.wolframalpha.com/input/?i=%28sqrt%28x%29%2B2%29%2B%2B%28x%2F3%2B%2B2%29>.
Foerster, Paul A. "Volume of a Solid by Plane Slicing." Calculus: Concepts and
Applications. Berkeley, CA: Key Curriculum, 1998. 242-51. Print.
Y=sqrt(x). Digital image. WolframAplha: Computational Knowledge Engine.
WolframAlpha, n.d. Web. 20 Feb. 2015.
<http://www.wolframalpha.com/input/?i=y=sqrt(x)&lk=4&num=1>.