application of integration

APPICATION OF INTEGRATIONAPPICATION OF INTEGRATION

SARDAR VALLABHBHAI PATEL INSTITAUTE OF TECHNOLOGY

GTU

CONTENTCONTENT

1.Volume by cylindrical shell2.Volume of solids of revolutions by

washers method

APPLICATIONS OF INTEGRATIONAPPLICATIONS OF INTEGRATION

Volumes by Cylindrical Shells

APPLICATIONS OF INTEGRATION

In this section, we will learn:How to apply the method of cylindrical shells

to find out the volume of a solid.

Let’s consider the problem of finding thevolume of the solid obtained by rotating aboutthe y-axis the region bounded by y = 2x2 - x3

and y = 0.

VOLUMES BY CYLINDRICAL SHELLS

If we slice perpendicular to the y-axis, we get a washer.

However, to compute the inner radius and the outer radius of the washer, we would have to solve the cubic equation y = 2x2 - x3 for x in terms of y.

That’s not easy.


Fortunately, there is a method—the method of cylindrical shells—that is easier to use in such a case.


The figure shows a cylindrical shell with inner radius r1, outer radius r2,

and height h.

CYLINDRICAL SHELLS METHOD

Its volume V is calculated by subtractingthe volume V1 of the inner cylinder from

the volume of the outer cylinder V2 .


Thus, we have:

2 1

2 22 1

2 22 1

2 1 2 1

2 12 1

( )( )( )

2 ( )2

V V V

r h r h

r r hr r r r hr r

h r r


Let ∆r = r2 – r1 (thickness of the shell) and

(average radius of the shell).

Then, this formula for the volume of a cylindrical shell becomes:

2V rh r

Formula 1

12 12r r r


The equation can be remembered as:

V = [circumference] [height] [thickness]


2V rh r

Now, let S be the solid obtained by rotating about the y-axis the region bounded by y = f(x) [where f(x) ≥ 0], y = 0, x = a and x = b, where b > a ≥ 0.


Divide the interval [a, b] into n subintervals [xi - 1, xi ] of equal width and let be

the midpoint of the i th subinterval.

xix


The rectangle with base [xi - 1, xi ] and height is rotated about the y-axis.

The result is a cylindrical shell with average radius , height , and thickness ∆x.

( )if x

( )if xix


Thus, by Formula 1, its volume iscalculated as follows:

(2 )[ ( )]i i iV x f x x


So, an approximation to the volume V of S is given by the sum of the volumes of these shells:

1 1

2 ( )n n

i i ii i

V V x f x x


The approximation appears to become betteras n →∞.

However, from the definition of an integral, we know that:

1

lim 2 ( ) 2 ( )n b

i i an i

x f x x x f x dx


Thus, the following appears plausible.

The volume of the solid obtained by rotating about the y-axis the region under the curve y = f(x) from a to b, is:

where 0 ≤ a < b

2 ( )b

aV xf x dx

Formula 2CYLINDRICAL SHELLS METHOD

The argument using cylindrical shellsmakes Formula 2 seem reasonable, but later we will be able to prove it.


Here’s the best way to remember the formula.

Think of a typical shell, cut and flattened, with radius x, circumference 2πx, height f(x), and thickness ∆x or dx:

2 ( )b

athicknesscircumference height

x f x dx


This type of reasoning will be helpful in other situations—such as when we rotate about lines other than the y-axis.


Find the volume of the solid obtained byrotating about the y-axis the regionbounded by y = 2x2 - x3 and y = 0.

Example 1CYLINDRICAL SHELLS METHOD

We see that a typical shell has radius x, circumference 2πx, and height f(x) = 2x2 - x3.


So, by the shell method, the volume is:

2 2 3

0

2 3 4

0

24 51 12 5 0

32 165 5

2 2

2 (2 )

2

2 8

V x x x dx

x x x dx

x x


It can be verified that the shell methodgives the same answer as slicing.

The figure shows a computer-generated picture of the solid whose volume we computed in the example.


Comparing the solution of Example 1 with the remarks at the beginning of the section, we see that the cylindrical shells method is much easier than the washer method for the problem.

We did not have to find the coordinates of the local maximum.

We did not have to solve the equation of the curve for x in terms of y.

NOTE

However, in other examples, the methods learned in Section 6.2 may be easier.

NOTE

Find the volume of the solid obtained by rotating about the y-axis the regionbetween y = x and y = x2.


The region and a typical shell are shown here.

We see that the shell has radius x, circumference 2πx, and height x - x2.


Thus, the volume of the solid is:

1 2

0

1 2 3

0

13 4

0

2

2

23 4 6

V x x x dx

x x dx

x x


As the following example shows, the shell method works just as well if we rotate about the x-axis.

We simply have to draw a diagram to identify the radius and height of a shell.


Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 1.

This problem was solved using disks in Example 2 in Section 6.2

y x


To use shells, we relabel the curve

as x = y2.

For rotation about the x-axis, we see that a typical shell has radius y, circumference 2πy, and height 1 - y2.

y x


So, the volume is:

In this problem, the disk method was simpler.

1 2

0

1 3

0

12 4

0

2 1

2 ( )

22 4 2

V y y dy

y y dy

y y


Find the volume of the solid obtained by rotating the region bounded by y = x - x2

and y = 0 about the line x = 2.


The figures show the region and a cylindricalshell formed by rotation about the line x = 2, which has radius 2 - x, circumference 2π(2 - x), and height x - x2.


So, the volume of the solid is:

0 2

1

0 3 2

1

143 2

0

2 2

2 3 2

24 2

V x x x dx

x x x dx

x x x


Volumes of Solids of Revolution: Washer Method

Washers!

Washers

Consider the area between two functions rotated about the axis f(x)

a b

g(x)

Washers

Consider the area between two functions rotated about the axis

Now we have a hollow solid

f(x)

a b

g(x)

Washers

Consider the area between two functions rotated about the axis

Now we have a hollow solidWe will sum the volumes of washers

f(x)

a b

g(x)

Washers

f(x)

a b

g(x) 2 2( ) ( )

b

a

V f x g x dx

Outer Function

InnerFunction

Find the volume of the solid formed by revolving the region bounded by y = (x) and y = x² over the interval [0, 1] about the x – axis.

Find the volume of the solid formed by revolving the region bounded by y = (x) and y = x² over the interval [0, 1] about the x – axis. 2 2([ ( )] [ ( )] )

b

a

V f x g x dx


b

a

V f x g x dx

1

0

222dxxxV


b

a

V f x g x dx

1

0

222dxxxV

1

0

4 dxxxV


b

a

V f x g x dx

1

0

222dxxxV

1

0

4 dxxxV 1

0

52

51

21

xxV


b

a

V f x g x dx

1

0

222dxxxV

1

0

4 dxxxV 1

0

52

51

21

xxV

103

application of integration

Engineering