Andrew Oughton
AP Calculus – Mr. Acre
10 February 2014
Area and Integrals of One Function
Given any function, it is possible to find the area below the curve by calculating the
definite integral. Integration calculates the area under the curve and within an interval by
dividing the area into many vertical strips, finding the area of those strips, and then adding all of
the areas together. In other words, the integral of a function (or the area below it) within a certain
interval of x-values (a and b) is the sum of the products of the y-coordinates (f(x)) and the change
in x (dx). The standard notation for integrals is:
𝑏
∫ 𝑓(𝑥)𝑑𝑥
𝑎
Figure 1. Integration
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One method of calculating the integral is the First Fundamental Theorem of Calculus.
The theorem states that for any function f and its indefinite integral F on the interval [a,b]:
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
𝑎
For example, say we need to calculate the area under the function 𝑦 =
𝑥
3
(F1) on the interval
from x = 3 to x = 10. The original integral would be:
10
∫
3
𝑥
𝑑𝑥
3
In order to use the Theorem, the function’s indefinite integral or antiderivative must be found:
∫
𝑥
𝑥2
=
=𝐹
3
6
Insert the values of a and b into the new function F and use the theorem to calculate the integral:
𝐹(𝑏) − 𝐹(𝑎) = 𝐹(10) − 𝐹(3) =
102 32
−
= 16.67 − 1.5 = 15.17
6
6
The final answer is 15.17. Therefore, the area under the graph of 𝑦 =
square units (u2).
2
𝑥
3
from 3 to 10 is 15.17
Another, significantly easier method to calculate definite integrals is to use a calculator or
computer. One of these tools is WolframAlpha, which, when given a function and interval, will
calculate the integral. When asked to find the definite integral of 𝑦 = √𝑥 (F2) from 4 to 8,
WolframAlpha will give the answer and its graph like this:
Figure 2. Integration from WolframAlpha
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Area and Integrals of Two Functions
Sometimes it is necessary to calculate the area between to functions. Situations with two
functions are similar to those with only one function in that they both have the same standard
notation and the symbols a, b, and dx mean the same things; however, in the case of two or more
functions, the height of the sections – f(x) originally – now becomes the difference in ycoordinates of the two functions. The standard notation is:
𝑏
∫ (𝑓(𝑥) − 𝑔(𝑥))𝑑𝑥
𝑎
Figure 3. Integration Between Two Functions
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Here is a graph of F1 (in red) and F2 (in blue):
Figure 4. Graph of F1 and F2
If we wanted to find the area between the two curves, first we would need to find where they
intersect:
√𝑥 =
𝑥
3
9𝑥 = 𝑥 2
0 = 𝑥 2 − 9𝑥 = 𝑥(𝑥 + 9), 𝑥 = 0,9
The graphs intersect at x = 0 and x = 9. These points become a and b in the integral, respectively.
Next, we must determine the height of the sections. As demonstrated in the following diagram,
the distance between the functions can be expressed as the difference between F2 and F1:
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Figure 5. Calculating Height
With this information, we can write out the integral and then solve:
9
𝑥
∫ (√𝑥 − ) 𝑑𝑥
3
0
2 3 1 2 |9
𝑥2 − 𝑥
3
6 |0
3
3
2
1
2
1
( (9)2 − (9)2 ) − ( (0)2 − (0)2 ) = 4.5 − 0 = 4.5
3
6
3
6
The final answer is 4.5, which means that the area between the curves in quadrant 1 is 4.5 u2.
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Volumes of Solids of Revolution
Solids of Revolution are 3-dimensional shapes formed by rotating a 2-dimensional area
around an axis. In the following image, the area of a 2-dimensional parabola bounded by the
graph and the x-axis in quadrant 1 is rotated around the y-axis, creating a 3-dimensional solid.
Figure 6. Solid of Revolution
7 http://mathdemos.org/mathdemos/washermethod/gallery/gallery.html
There are three methods to find the volume of a solid of revolution: the Disk method, the
Ring method, and the Shell method. Each method utilizes multiple cylindrical cross sections
taken from the solid to estimate the solid’s total volume. Which method is used depends on
which axis the graph is being revolved around and in what direction the cross sections are taken.
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Method 1: The Disk Method
Figure 8. The Disk Method
9 http://calculus2010.wikidot.com/volumes
The Disk method is used when one of the bounds of the area being revolved is the axis of
revolution and the cross sections are taken perpendicular to the axis of revolution. In the picture
above, the shape is revolved around the x-axis and the cross sections are taken perpendicular to
the x-axis (dx).
The volume of the solid is found by adding the volumes of many cylindrical cross
sections taken out of the solid. Each individual volume can be calculated using the formula for
the volume of a cylinder:
𝑉 = 𝜋𝑟 2 ℎ
In the case of the picture above, the radius of the cylinder is the y-coordinate of the
function (f(x)) and the height (or thickness) is the change in x or dx. The new volume formula of
one disk, then, is:
𝑉 = 𝜋 ∗ 𝑓(𝑥)2 𝑑𝑥
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In order to find the volume of the entire solid, all of the individual cylinders must be
added. This can be done by integrating the volume formula within the bounds of the area. For the
example above, this would be:
3
∫ 𝜋 ∗ 𝑓(𝑥)2 𝑑𝑥
−3
If the axis of revolution is vertical instead, then f(x) in the integral would be the xcoordinate and the dx would be dy. Otherwise, the integral is set up the same way.
Method 2: The Ring Method
Figure 10. The Ring Method
11 http://calculus2010.wikidot.com/volumes
When the cuts are taken perpendicular to the axis of rotation but there is a gap between
the area and the axis of rotation, the Ring method must be used to find the volume of the solid. In
the example picture above, the yellow area is being rotated around the x-axis and the cuts are
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taken perpendicular to the x-axis, but there is a gap between the shape and the axis, meaning that
the disk formed by rotation has a hole in it.
Figure 12. F1 and F2, and y = -2
13
http://graphsketch.com/
The volume of the ring can be thought of as a large disk with a smaller disk cut out of its
center. The ring’s volume can then be expressed as the volume of the larger disk minus the
volume of the smaller disk. For the graph above, if F1(x) is the function in red and F2(x) is the
function in blue, and the area between them is being rotated around the line y = -2, in green:
𝑉 = (𝜋 ∗ (𝐹2(𝑥) + 2)2 𝑑𝑥) − (𝜋 ∗ (𝐹1(𝑥) + 2)2 𝑑𝑥)
Which can be simplified to:
𝑉 = 𝜋 ∗ ((𝐹2(𝑥) + 2)2 − (𝐹1(𝑥) + 2)2 )𝑑𝑥
Two was added to each radius because the axis of rotation was two units below the xaxis, so the radius is no longer just the y-coordinate, but two plus the y-coordinate.
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In order to calculate the volume of the entire solid, the integral of the volume formula
bound by a and b, which are 0 and 9 respectively, must be taken. In the case of the example
above:
9
∫ 𝜋 ∗ ((𝐹2(𝑥) + 2)2 − (𝐹1(𝑥) + 2)2 )𝑑𝑥 ≈ 98.960
0
The result from integration was 240, which means the total volume of the rotated solid
formed when the area between the functions F1 and F2 is rotated around the line y = -2 is
98.960u3.
Just like with the Disk method, if the axis of rotation is vertical rather than horizontal and
the cuts are dy cuts, the integral becomes:
𝑏
∫ 𝜋 ∗ (𝑓(𝑦)2 − 𝑔(𝑦)2 )𝑑𝑦
𝑎
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Method 3: The Shell Method
Figure 14. The Shell Method
15 http://www.mathdemos.org/mathdemos/shellmethod/gallery/gallery.html
The Shell method is used when the axis of rotation and the direction of the cross sections
are parallel. In the example picture, the original function, represented in purple, is rotated around
the y-axis and the cross sections are taken parallel to the y-axis (dx cuts). Each individual
volume, instead of being calculated as disks or rings, is calculated as a cylindrical shell. The
volume formula for a cylindrical shell is the circumference of the circular base, times the height
of the cylinder, times the cylinder’s thickness or:
𝑉 = 2𝜋𝑟ℎ𝑑𝑟
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For the example image above, the radius (r) of each hollow cylinder is the x-coordinate.
The height (h) is the y-coordinate or f(x). Finally, the thickness (dr) is the change in x, or dx. The
volume formula for each cylindrical shell is then:
𝑉 = 2𝜋 ∗ 𝑥 ∗ 𝑓(𝑥)𝑑𝑥
As with the other two methods, to find the volume of the entire solid, the integral of the
volume formula must be taken – this adds together all of the possible hollow cylinders that can
be made from the function to find the total volume. This integral is:
𝑏
𝑏
∫ 2𝜋 ∗ 𝑥 ∗ 𝑓(𝑥)𝑑𝑥 = 2𝜋 ∫ 𝑥 ∗ 𝑓(𝑥)𝑑𝑥
𝑎
𝑎
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Finding Volume with Cross-Sections
Figure 16. Volume with Square Cross Sections
17 http://www.education.com/study-help/article/solids-cross-sections/
If the formula to find the volume of a cross-section of a solid is known and stays the same
throughout the solid, the volume of the solid can be calculated using integrals. In the example
picture above, the 2-dimensional base of the solid is a triangle, and each cross section is a square.
By finding the volume of each cross section and adding them together, the total volume of the
pyramidal solid can be found.
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Figure 18. Graph of F1 and F2
19 http://graphsketch.com/
In the graph above, which contains functions F1 and F2, a solid is formed from cross
sections perpendicular to the x-axis of isosceles right triangles in the area between both functions
with one leg as the base. If we want to find the volume of the solid, we must first find the volume
of a cross section.
Each cross-section is an isosceles triangle with one leg as the base. The formula for the
volume of each cross-section is half of the base times the height time the width, or:
1
𝑉 = 𝑏∗ℎ∗𝑤
2
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If the leg is the distance between the functions, the length of the leg must be 𝐹2(𝑥) − 𝐹1(𝑥). In
an isosceles right triangle, the other leg is the height of the cross-section and is also equal to the
length of the other base. Also, because the cross-sections are perpendicular to the x-axis, the
width must be dx. The new volume formula is:
1
1
𝑥 2
2
𝑉 = (𝐹2(𝑥) − 𝐹1(𝑥)) 𝑑𝑥 = (√𝑥 − ) 𝑑𝑥
2
2
3
Integrating the formula from 0 to 9, the bounds of the area, will give the volume of the
total solid:
9
∫
0
1
𝑥 2
1 9
𝑥 2
(√𝑥 − ) 𝑑𝑥 = ∫ (√𝑥 − ) 𝑑𝑥 = 1.35
2
3
2 0
3
The final result is 1.35, which means that a solid formed from isosceles right triangles with one
leg as the length between F1 and F2 has a volume of 1.35u3.
This process can be repeated for any area and any shape cross-section as long as there is
enough information about the area that forms the base. Without that information, there is not
enough to find the volume formula of the cross-sections and therefore the volume of the whole
solid.
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