Applications of

IntegrationHanna Kim & Agatha Wuh

Block A

Disk Method IntroDisk Method is in many

ways similar to cutting a rollcake

into infinitely thin pieces and

adding all of them together

to find out the volume of the whole.

Delving into Disk Method

•Disk = right cylinder

•Volume of disk= (area of disk) * (width of disk)=πr2w

•Integration is simply an addition of infinitesimally thin disks in a given interval.

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Demonstration of how the whole volume is computed by the addition

of smaller ones

Disk Method Integration

Horizontal Axis of Revolution

Vertical Axis of Revolution

V =π [R(x)]2a

b

∫ dx

V =π [R(y)]2c

d

∫ dy

πr2=Area of Circle

y =sinx+ 3

π(sinx+ 3)2 dx−π2

3π2

∫

fnInt((π(sin(x) + 3)2 ),x,−π /2, (3π ) / 2 ≈187.522 units3

The area under the curve y is rotated about the x-axis in the interval [-π/2, 3π/2]. Find its volume.

Disk Method Example

Washer Method Intro

Washer’s method is

an extension of disk method. However, the

difference is that there is a hole in

the middle just like a bamboo or a

collection of CDs!

•Washer=a disk with a hole

•Using this method is to find out an area of washer by subtracting the smaller circle from the bigger circle and then adding all the washers up.

Delving into Washer Method

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Washer Method Integration

V =π [R(x)]2 −[r(x)a

b

∫ ]2dx

Washer Method Example

The region bounded by y=-5(x-3)2+8 and y=sin x +2 is revolved around the x-axis. Find its volume.

y =−5(x−3)2 + 8y=sinx+ 2

Find points of intersection

x =1.991 and x=4.171

1. Graph the two equations.2. Press 2nd Trace or Calc3. Press 5: intersect4. Identify the two different graphs.5. Done!

Washer Method Example

π R2 − r2

a

b

∫ dx

π [(1.991

4.171

∫ − 5(x − 3)2 + 8)2 − (sin x + 2)2 ]dx

fnInt((−5(x − 3)2 + 8)2 − (sin(x) + 2)2 , x,1.991,4.171)

≈ 75.713

Ans *π ≈ 237.860 units3

Outer circleInner circle

Shell Method IntroShell Method is like

“unrolling” a toilet paper and adding up the

infinitely thin layers of paper.

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Delving into Shell Method

Volume of shell= volume of cylinder-volume of hole=2πrhw

2π(average radius)(height)(thickness)

2πrh=circumference*height

=Area of cylinder without the top and bottom circles of the cylinder

Shell Method Integration

V =2π p(y)h(y)dyc

d

∫ V =2π p(x)h(x)dxa

b

∫

Shell Method ExampleThe region bounded by the curve , the x-axis,

and the line x=9 is revolved about the x-axis to generate a solid. Find the volume of the solid.

y = x

(9,3)y = x

x=y2

2π(y)(9 −y2 )dy0

3

∫

2π 9y−y3 dy0

3

∫

2π[9y2

2−

y4

4 0

3]

2π[9 * 32

2−34

4]

81π2

radius heightThe integral is in terms

of y because y is parallel to the axis of

rotation.

Shell Method Example

y = x

x=y2

2π(y)(9 −y2 )dy0

3

∫

2π 9y−y3 dy0

3

∫

2π[9y2

2−

y4

4 0

3]

2π[9 * 32

2−34

4]

81π2

Constant multiple rule

Evaluate from 0 to 3

Answerunits3

Section Method Intro

The section method is like

cutting a loaf of bread into

different slices and adding

those slices up.

•Utilizing simple area equation of the cross section.

•Formula is applicable to any cross section shape

•Most common cross sections are squares, rectangles, triangles, semicircles, trapezoids

Delving into Section Method

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Section Method Integration

V = A(x)dxa

b

∫

Cross section taken perpendicular to the x-

axis

Cross section taken perpendicular to the y-

axis

V = A(y)dya

b

∫

Section Method Example

The solid lies between planes perpendicular to the y-axis at y=0 and y=2. The cross sections perpendicular

to the y-axis are half circles with diameters ranging from the y-axis to the parabola

x = 5y2

d= 5y2

r =52

y2

The integral is in terms of y because the

cross sections are

perpendicular to the y-axis.

Section Method Example

1

2π(

52

y2 )2 dy0

2

∫12π(

54)y4 dy

0

2

∫58π y4 dy

0

2

∫58π[

y5

5 0

2]

58π[

25

5−05

5]

58π[

325]

4π

1

2π(

52

y2 )2 dy0

2

∫12π(

54)y4 dy

0

2

∫58π y4 dy

0

2

∫58π[

y5

5 0

2]

58π[

25

5−05

5]

58π[

325]

4πConstant multiple rule

1/2 πr2=half circleEvaluate

Simplify

units3 Answer

Shell Method VS.

Disk Method

How Can We Distinguish?•For disk method, the

rectangle (disk) is always perpendicular to the axis of revolution

•For shell method, the rectangle is always parallel to the axis of revolution.

Works Cited

•http://library.thinkquest.org/3616/Calc/S3/TSM.html

•http://mathdemos.gcsu.edu/mathdemos/shellmethod/2curvesshells.gif

•http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html

•http://mathdemos.gcsu.edu/mathdemos/washermethod/