Download CHAPTER SIX: APPLICATIONS OF THE INTEGRAL

APPLICATIONS OF THE
INTEGRAL
Released by Eman Salem Al-Aidarous
October, 2010
AREA
In this presentation,
 we introduced integration in terms of area
between the curve and the x-axis.
 We will discuss area between two curves and
methods.
Calculus Strategies: Integration

The definite integral is
an accumulation of
products … that is the
sum of products of two
quantities, so definite
integrals can be thought
of as measurements of
areas.
f x = 0.1 25-x2
4
2
-5
5
3
Calculus Strategies: Integration

In any application of integration (such as areas under
a curve, volumes, arclength, work, distances, or total
costs), there is a three step strategy:



Cut the <area, volume, arclength, work, etc> into small
pieces.
Code the quantity to be measured on a representative small
piece, because we understand the geometry of the small
parts.
Recombine the parts (with sums / definite integrals).
4
Calculus Strategies: Integration

Step 1 (Cut the desired
result into small pieces.)
f x = 0.1 25-x2
4
2
f x = 0.1 25-x2
2
-5
-5
5
5
-2
5
Calculus Strategies: Integration

Step 2 (Code the quantity to be measured on
a representative small piece)

It looks like this:
f x = 0.1 25-x2
4
2

dA = y dx
-5

5
The width (x) is cut into infinitesimally small
parts, and the height (y) depends on the
function under which the area is to be
measured.
6
Calculus Strategies: Integration

Step 3 (Recombine the parts with sums /
definite integrals )
f x = 0.1 25-x2
4

dA = y dx
2
-5

5
Adding up all of these simpler parts becomes
b
A   ydx
a
5

A   (25  x 2 )dx
0
7
AREA BETWEEN TWO CURVES



The area between two curves is really simple if you really look
at it. Let’s consider f(x)=6-x2. Also consider g(x)=x.
Find the area between these two curves.
Here is a graphical view of the scenario. (Magenta region)
ALWAYS DRAW A PICTURE TO SEE HOW THE
REGION WILL LOOK LIKE!!!!!!! 
STEP 1: FIND THE POINTS OF
INTERSECTION



In order to see where the
region “begins” and
“ends,” we must find the
x limits to see where we
can actually compute
area.
Solve for the x limits
The two limits of
integration are x=2 and
x=-3
6 x  x
2
0  x  x6
2
0   x  2  x  3
x2
x  3
STEP 2


Compute the area of both functions.
Subtract the areas.
2
1 3
28
55

36  x dx  6 x  3 x  3  3  9  3
2
1 22
9
5
3xdx  2 x 3  2  2   2
55  5  125
  
3  2
6
2
2
 
THIS IS OUR AREA BETWEEN THE TWO CURVES
GEOMETRICALLY


This is a shaded region style problem. Let the maroon be the region
between the parabola and the x axis. with the limits of -3 and 2. Let the
light blue be the region between the line and the x axis with the same
limits. Let the purple represent overlapping regions.
We must subtract the two areas of the regions (that is Af(x)-Ag(x)) to get
rid of the overlapping areas (i.e. purple region).
FORMULA FOR FINDING AREA
BETWEEN TWO CURVES

If f(x)>g(x), then…
A    f ( x)  g ( x)dx
b
a
Top Function
b

Bottom Function

A   ytop  ybottom dx
a
If f(x)>g(x)… top vs. bottom
function???

Very simple..

Since the
parabola, f(x), is
on top of the
line, g(x),
therefore,
f(x)>g(x).
TOP
BOTTOM
AREA BETWEEN TWO CURVES




We just tackled one of the three types of area
between curves problems. They are
1) Area between two y(x) functions. (We just
did)
2) Area between a y(x) and x(y)
3) Area between two x(y) functions.
AREA BY SUBDIVIDING



Sometimes, it is necessary to split the region
up into two parts before integrating. This
situation often occurs when you have y as a
function of x and x as a function of y.
Consider this example: Find the region
bounded by y=½x and x=-y2+8.
First thing you do is to draw a picture to see
how the region will look like. Then solve for
their intersecting points.
DRAW A PICTURE!!!

It helps to draw a picture to get an understanding on
what is going on. The red line is y=½x and the blue
parabola is x=-y2+8.
THE INTERSECTING POINTS
ARE (-8,-4) and (4,2)
1
1 2
2
y  x y  x
2
4
x  8  y 2
y2  8  x
1 
4 x 2   48  x 
4 
x 2  32  4 x
x 2  4 x  32  0
x  8x  4  0
x  8
y  4
x4
y2
INTEGRATING


It is very important to
let all the functions be 1
in the form y(x).y  2 x
Therefore, solve for y. y  8  x
Using the formula given 4 1
x  8  x dx

previously for finding 8 2
4
1
2
the area between curves, x 2  8  x 3 / 2    4    80   76


4

3
3  3 3
 8
apply that formula using
-8 and 4 as your limits.
WHAT AREA DID WE JUST
DO???



Notice how much area we did accumulate using the integral.
Also consider how much area we have left to take into
account.
The purple is the area we jus did. (76/3) The green is the area
we have left to do.
THE GREEN AREA

Since we know that the area of the green is the
area of that sideways parabola, and since we
also know that the area above the x-axis from
[4,8] equal to the area below the x-axis. So in
effect, we can double the area of that to take
into account the top and the bottom of the xaxis.
CALCULATION OF AREA

Notice the 2 outside the
integral. We have to
double the area.
2
8
4

Fundamental theorem.

Add the two areas.
THE TOTAL AREA
OF THE REGION IS
36!!

8  x dx
8
32 
 4

3/ 2 
 3 8  x     0  3 

4 

76 32

 36
3
3
INTEGRATING A FUNCTION
WITH RESPECT TO Y.

At times, we may have to curves that there is
no “top” or “bottom” y, rather there are left
and right curves. By definition, these curves
would not be functions of x, since they would
fail to honor the vertical line test (the test that
determines if a relation is actually a function).
But these functions, on the hand, seem simpler
if they were in terms of y.
INTEGRATING WITH RESPECT
TO Y.

To calculate an area, using the right being x=f(y) and
the left one being x=g(y) function, between limits of
y (c and d)…then…
A    f ( y)  g ( y)dy
d
c
d

OR

A   xright  xleft dy
c
EXAMPLE




Let’s do the previous problem using this method of integrating with respect
to y.
The graphs are y=½x and x=-y2+8.
First, always always always!! Look at a picture of this! This will help you!
The red is the line. The blue is the parabola. The purple is the region.
RIGHT
LEFT
SOLVE FOR INTERSECTING
POINTS
1
1
x  y2  x2
2
4
x  8  y 2
y
y2  8  x
1 2
4 x   48  x 
4 
x 2  32  4 x
x 2  4 x  32  0
x  8x  4  0
x  8
y  4
x4
y2
SAME SLIDE AS BEFORE, HOWEVER, CONSIDER
THAT THE Y VALUES ARE NOW MORE
RELEVANT, SINCE WE ARE INTEGRATING WITH
RESPECT TO Y!!!!
THIS MEANS THE INTEGRAL WILL HAVE THE
LIMITS BETWEEN y=-4 AND y=2!!!
SETTING UP THE INTEGRAL

First solve everything for x before putting the f(y)
and g(y) inside. You will get x=2y AND x=8-y2.
A    f ( y )  g ( y )dy
d
c
2


A   8  y 2  2 ydy
4
2
A    y 2  2 y  8dy
4
2
 1  3

2
  3  y  y  8 y 


 4
 28   80  108
 36
    
3
 3   3
AREA BETWEEN TWO CURVES



In reality, there is no set formula to finding area
between curves. It’s important to know what to do
and how you go about doing it. For example, you
noticed that integrating a function in terms of y was
simpler than taking the region, cutting it, and finding
areas that way.
But sometimes, you will see that integrating with
respect to y can be a hassle. Sometimes, integrating
with x would be better.
But in either case, the general formula stays the same:
top/right – bottom/left. (for x and y respectively)
VOLUME OF A SOLID OF
REVOLUTION



FOR AP CALCULUS AB STUDENTS THIS IS
YOUR FINAL TOPIC
FOR CALCULUS I: YOU ARE COMPLETED FOR
THE COURSE. YOU ARE WELCOMED TO
STAY, BUT YOU CAN LEAVE IF YOU WISH.
TAKE CARE! BEST WISHES! HARE KRSNA!
FOR CALCULUS II STUDENTS: HARI BOL!
WELCOME! THIS IS YOUR FIRST TOPIC FOR
THIS CLASS!!!!
VOLUME


Imagine if you were to take a region between
f(x), the limits of integration, and the x-axis,
and you were to swing that region about the xaxis.
Take your hand and try that. Put your hand on
the region, and turn your hand such that you
rotate about the x-axis (2 pi).
SEE THE BEFORE AND AFTER
PICTURES!!!
CLARIFICATION



What we did was we took the region under the
line from x=0 to x=7 and rotated that region
360 about the x-axis.
You will notice that a solid is formed. That
solid looks like a cone in this example.
How do we find the volume of this cone using
calculus?
IF THE WIDTH

If the width was infinitesimally small as dx, then the
area would represented as an integral.
V ( x) 

b
A( x) dx
a
CUTTING UP THE SOLID

Imagine if we were to cut up the solid to infinity
slices! That would that the thickness of the slice
would be virtually 0. But let’s take out one of these
slices …
dx
THE SLICE

The cross section of such volume is a circle.
The radius of the circle would merely be f(x)
value, since the center of the circle is the xaxis. Look at the 2-D and the 3-D illustration.
The radius of the cirlce
is f(x).
r=f(x)
x axis
AREA OF A CIRCLE?






Let’s say that the line was the function
f(x)= ½ x
Remember that the integral of the area of the cross
sections from a to b is the volume? Also remember
that f(x) is the radius?
If the circle is the cross-section for the solid, then
what is the area of a circle?
C’mon! Don’t tell me you are in calculus and don’t
know the area of a circle!
A=pr2 !






If the radius and the
function are equal, so are
their squares
f(x)=½x
Apply the squared radius
into the area of circle
expression
Volume-as-an-integral
expression
Limits of the original
problem were x=0 and x=7.
Evaluation of the integral.
The volume is found
r   f ( x) 
2
2
1 
r   x
2 
1 2
2
r  x
4
π 2
A x
4
2
2
b
V ( x)   A( x)dx
a
V ( x)  π 
7
0
x2
dx
4
7
243π
 π 3
V ( x)   x  
12
12  0
RECALL

We took a region under f(x)=½x from x=0 to
x=7 and rotated it around the x- axis to form a
solid. We took this solid and cut it up to
infinite slices. We took one of those slices and
examined that the cross-section is a circle
(most commonly called a disk). We found the
area formula for the disk. A=pr2. Since r=f(x),
we put it into the formula. The volume formula
says that we take the integral from 0 to 7, for
the [f(x)]2. That result yields into the volume.
DISK METHOD FORMULA



If the cross sections are
circles, and if region are is
being rotated around the xaxis, then the formula for
finding volume is:
If the region between g(y)
and x -axis is being rotated
about the y-axis, with disk
cross section, then the area
formula is this.
Don’t forget to put the p in
there.
V ( x)  π   f ( x) dx
b
2
a
V ( y)  π  g ( y) dy
d
c
2
DISK METHOD


Remember how we had instances that we were
forced to integrate with respect to y?
If that is the case, solve for y, use the y-values
for the limits of integration, and use the same
formula.
WASHER METHOD

Find the volume of the region between y2=x,
and y=x3, which intersect at (0,0) and (1,1),
revolved about the x–axis.

First, DRAW A PICTURE
PICTURES
THE SOLID


Notice that the solid now has a “hole” inside it
(very light blue). The edges of this solid
(darker light blue) (is now created by two
functions.
Since we are revolving this solid around the xaxis, we must solve everything in terms of x.
SHELLS FORMULA


If we solve for x, then we will get √x. (square root)
and x3.
If you look at the cross section, you will get
something similar but somewhat different from a
disk. In order to take into account, the hole in the
bottom, we have to include a hole inside the disk.
Therefore, the slice will look like this.
g(x) BOTTOM FUNCTION
f(x) TOP FUNCTION
THE AREA OF THE ORANGE

If you wanted to find the area of the orange portion,
then find the area of the entire circle and subtract the
area of the inner cirlce. A typical shaded region
problem!  Capital R means outer radius, and lower
case r means “inner radius.”

A π R r
2
2

VOLUME

Having infinite “doughnuts” that are very thin, you
can use the integral to find the volume. This is called
WASHER’S FORMULA.

V ( y)  π  R
b

 r dy
V ( x)  π  R  r dx
2
a
d
c
2
2
2
R and r


If you also consider it, the outer radius is the
top function while the inner radius is the
bottom function, since the bottom function
borders the hole in the center of the
“doughnut.”
Let’s call this “doughnut” a washer from now
on!!!!
FINDING THE VOLUME


Wahser’s formula
DON’T FORGET THE V ( x)  π b R 2  r 2 dx

a
p!
1
V ( x)  π 
0


 x   x  dx
2
3 2
Original limits and top V ( x)  π 1 x  x 6 dx
0
and bottom functions
2
7 1
are included.
x
x 
 1 1  5π
V ( x )  π     π   
The
final
volume
 2 7  14
 2 7 0
evaluated.
INTEGRATING w/ RESPECT TO Y.




Let’s take the previous problem and rotate it
about the y-axis.
That means, we will have R as right function
and r as the left function
Therefore: x=y2 and x=y⅓.
ALWAYS DRAW A BEFORE/AFTER
PICTURE!!!
PICTURES
VOLUME rotated about y axis


V ( y )  π   y    y  dy
V ( y )  π   y    y dy
d
V ( y )  π  R  r dy
2
2
c
1
1/ 3 2
2 2
0
1
2/3
4
0
1
 3 5/3 x 
3 1 2
V ( y )  π  x    π    π
5 0
5 5 5
5
5
LINE ROTATIONS




Suppose we take the previous example and
rotated it about the following lines
A) x=-1
B) y=-1
Find both volumes.
DRAW A PICTURE

Draw both before/after diagrams for the x=-1
rotation and y=-1 rotation.
PICTURES for rotation about
x=-1
PICTURES for rotation about
y=-1
THE SHIFT!!!



Look at the reference line (reference line is the
line you are rotating the region about.)
Actually draw out the radii. For the x=-1
rotation, you will need to add 1 to both
functions to take into account.
General Rule: If rotation around a line x=k,
then f(x)-k is the big radius and g(x)-k is the
small radius.
THE RADII


Therefore, the radii is 1+y2 and 1+y1/3.
Since by this time, you are getting the
fundamental theorem of calculus down pretty
well (or at least you should be), I will stop
showing the plugging in of numbers. I will
show the integral, substitution, actual
integration and answer.
INTEGRATION


V ( y )  π  1  y   1  y  dy
V ( y )  π  1  2 y  y   1  2 y
d
V ( y )  π  R  r dy
2
2
c
1
1/ 3 2
2 2
0
1
0
1/ 3
2/3
2

 y 4 dy
1

6 4/3 3 5/3  
2 3 y 
V ( y )  π  y  y  y    y  y  
4
5
3
5  0
 

 23 13  37
V ( y )  π   
π
 20 15  30
5
ROTATING about y=-1



Remember! If you have y=k as line of rotation,
then you will f(x)-k and g(x)-k
Therefore, you will have f(x)=x1/2 +1 and
g(x)=x3+1
Using the washers formula, let’s find the
volume.
VOLUME
b


V ( x)  π  R 2  r 2 dx
a
    dx
V ( x)  π  1  2 x  x  1  2 x  x dx
1
V ( x)  π  1  x  1  x
2
3 2
0
1
3
6
0
1

4 3/ 2 x  
x
x 
V ( x)  π  x  x     x   
3
2 
2 7  0

4 1 1 1 π
V ( x )  π     
3 2 2 7 3
2
4
7
VOLUME BY SHELLS


If we wanted revolve a region to form a solid,
we could use really thin cylinders, find the
area for all of these cylinders and thus, get the
volume.
It is merely the sum of the surface area of each
cylinder. The formula of the surface area of a
cylinder is 2prh. If we make the thickness as
thick as dx, then the shells formula would like
this.
SHELLS FORMULA
V ( y )  2 π  Rxdx
V ( x)  2 π  Rydy
r
h
For rotation over the y -axis
ARC LENGTH AND SURFACE
AREA



This topic requires very strong algebra skills.
We will introduce arc length and surface area
of a solid of revolution, since their formulas
are very much similar.
First, we must discuss “ds.”
WHAT IS “ds?”
PYTHAGOREAN THEOREM



Remember the Pythagorean theorem?
x2+y2=r2
Let’s call they hypotenuse as “s” instead of
“r.”
s
y
x
SIMILARLY…




Works with differentials!
ds2=dx2+dy2
Same concept applies
You could rewrite it as ds2=[1+(dy/dx)2]dx2
ds
dy
dx
FINDING DISTANCE
Ds
Ds
Dx
Dy
Dx
Ds
Dy
Dx
Dy
DISTANCE




Notice if you make right triangles with equal ∆x, the
hypotenuse will be really close to the graph line. The
hypotenuse would be ∆s.
The sum of ∆s will give you the distance or the length
of the line.
If we took infinitesimal right triangles such that ∆x
would become the width, dx. Therefore, ∆s will
become ds.
The sum of these very minute lines or “dots” would
become the length of the line.
REIMANN SUM AND INTEGRAL
FOR ARC LENGTH

Definition for ds.

Reimann sum to find the
distance of the line


THE
ARC
FORMULA with
x.
THE
ARC
FORMULA with
y.
2
LEGNTH
respect to
 dy 
ds  1    dx
 dx 
n
n
i 1
i 1
 Ds  

b
a
LENGTH
respect to
ds  
b
a
Dx  Dy Dx
2
 dy 
1    dx
 dx 
2

d
c
ds  
d
c
 dx 
1    dy
 dy 
COMPUTATING ARC LENGTH



Due to the radical sign in the ds expression,
integration becomes really difficult.
In the later chapters, we will discuss
integration of such integrals.
We will only do very simple functions that we
can integrate with the knowledge we have
now.
FROM THIS FORMULA

You can get surface
area.
Asurface( x)  
b
a
2
 dy 
y 1    dx
 dx 
2
Asurface( y )  
d
c
 dx 
x 1    dy
 dy 
SAMPLE PROBLEM


Let’s say you are given the great honor to make
garland for the Deities. Of course, you don’t want
them to be so short that the garlands become a
necklace. But then again, you don’t want to make
them as long as they would touch the floor! In other
words, you gotta make ‘em just right!
Make this a bonafide problem as much as possible.
If we were to “break the loop and lie it along the
“tilak graph” (y = (2/3)x3/2), you’ll see that the graph
starts from x=3 to x = 15. Find the length of this
garland. 
SAMPLE PROBLEM
15

3
Ldx 
15

3
2
 dy 
1 
 dx
 dx 

y  x3/ 2
dy

dx

x
2
 dy 

 x
 dx 
15

3
Ldx 

15

3
1  x dx
u  1 x

du  1dx
15

3
Ldx 
16

16
4
u du
2
2
112
 2 3/ 2 




u

64

8

56

3

3
3
3

4

You guessed it! Arc length
formula!
You differentiate the function
and square the derivative as
shown.
Put the formula to some
action!!
Oops!
Don’t
forget
usubstitution. If u=1+x, then
du/dx = 1, therefore du = dx.
Using
the
fundamental
theorem of calculus along with
the u-substitution, evaluate the
integral! 
SURFACE AREA PROBLEM




Remember in the Mahabharata, the five Pandavas
needed to be in disguise during their thirteenth year
of exile! Arjuna is gonna be Brhanalla, and he needs a
pakhwaj drum.
You are his personal assistant who makes these drum.
It’s almost cylindrical but let’s say that is 100%
cylindrical.
Theoretically, the pakhawaj resembles the solid
formed by rotating the region over the x-axis. The
region is defined as being bound between the line
y=3, x-axis, y axis and x=10.
Find the surface area of this pakhawaj.
GRAPH, BEFORE, AFTER

Always know what region
you are talking about.
Draw a graph!! It helps!

Eh.. I know.. Not the best
pic… but you get the idea
of how it looks like when
rotated!
FINDING THE SURFACE AREA
Asurface( x)  
b
a
2
 dy 
y 1    dx
 dx 

y3
dy
0
dx

10
Asurface   3 1dx

0
10
Asurface   3dx

0
10

0
3dx  3 x 0  30  0  30
10

Remember! Formula first!
Don’t forget! We rotated
about the x-axis, so we use
the A(x) formula.
The function is y=3. dy/dx =
0.
The integral becomes a
piece of cake
Simply apply fundamental
theorem of calculus.
Don’t forget. Area = units
squared!

A   x

b
A   ytop  ybottom dx

Area between 2
curves

(all)
V ( y )  π  g ( y ) dy

Disk method
b

(AB II)

Washer’s method

(AB II)

Cylindrical Shells

(BC II)

Arc Length

(BC II)

Surface Area

(BC II)
a
d
c
right

 xleft dy
V ( x)  π   f ( x) dx
b
2
a
d
2
c

V ( y )  π  R

 r dy
V ( x)  π  R 2  r 2 dx
a
d
2
c
2
V ( y )  2 π  Rxdx
V ( x)  2 π  Rydy

b

d
a
ds  
b
a
2
 dy 
1    dx
 dx 
2
c
ds  
d
c
 dx 
1    dy
 dy 
Asurface( x)  
b
a
2
 dy 
y 1    dx
 dx 
2
Asurface( y )  
d
c
 dx 
x 1    dy
 dy 
SUMMARY




When doing these problems with area, volume,
arc length, or surface area, always draw a
picture of the all graphs and conditions
involved.
Then draw a before/after scenario.
Write the equations necessary to solve the
problem.
Check with a calculator if available 
CONCLUSION



We are about to reach a point in calculus where
simply memorizing equations and plugging numbers
will not be useful all on its own.
Calculus requires critical thinking. The secret behind
math is planning. How do you plan to solve a
problem? With method would work the best and
efficiently? These questions will come up soon.
Never forget the important rules of integration. The
next few chapters are strictly devoted to integration.