Calculus: Volume by Integration - Solids of Revolution, Slides of Differential and Integral Calculus

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TOPIC

APPLICATIONS

VOLUME BY INTEGRATION

• define what a solid of revolution is
• decide which method will best determine the

volume of the solid

• apply the different integration formulas.

OBJECTIVE

This method is used when the element (representative

strip) is perpendicular to and touching the axis. Meaning,

the axis is part of the boundary of the plane area. When

the strip is revolved about the axis of rotation a DISK is

generated.

A. DISK METHOD: V = r2h

h = dx

y

dx

x = a

f(x) - 0

x = b

y = f(x)

x

= r

The solid formed by revolving the strip is a cylinder whose volume is

V   r^2 h V  f x  dx

2  ( ) 0

To find the volume of the entire solid (^)    

b

a

V f x dx

2  ( )

Example: Find the volume of the solid generated by revolving the region bounded by the line y = 6 – 2x and the coordinate axes about the y-axis.

r =x

( x , y ) h = dy

( 0 , 6 )

x

y

0

(3,0)

By horizontal stripping, the elements are perpendicular to and touches the axis of revolution, thus we use the disk Method.

We use V   r^2 h to find the area of the strip.

6

0

3

6

0

2

6

0

2

6

0

2

2

6 y

V

6 y dy

V

6 y dy

V

6 y

V x dy ,if y -2x 6; x

dV x dy

^ 

V 18 cu. units

36 6 12

V

6 6 6 0 12

V 3 3

 



  

  

  

( x1 , y1 )

( x2 , y2 )

x = a x = b

dx

h = dx

y1 = g(x) (^) y2 = f ( x )

 

^ ^ 

   

  b V a y y dx

dV y y dx 2 2

2 1

2 2

2 1



 Since^ y^1^^  f ( x ) y 2 (^)  g ( x )

 ^ ^ ^  (^) 

 (^) 

   

b a

2 2 V g x f x dx

and

r

R

Figure 6.2.15 (p. 427)

. x^2 ^4 y x ^1 ^0

Example: Find the volume of solid generated by revolving the second quadrant region bounded by the curve about.

R =1- x

h =d y

( 0 , 4 )

x 0

x2 = 4-y x- 1=

r =1- 0 = 1

y



(-2, (^0)  )

By horizontal stripping, the elements are perpendicular but not touching the axis of revolution, thus we use the Ring or Washer Method.

V  ^ R^2  r^2 ^ h

 

2

R  1 ( 4  y ) ^ (^1 ^4 y^ )^2 r ^1

dV  ^ ^ ( 1  4  y)^2  12 dy

V ( 1 2 4 y 4 y 1 ) dy

4

0

      

 ( 4 y) 2 4 ydy

4

0

    

0

4 3 / 2 2 (^4 y^ ) 2

( 4 y )^23 /^2  

(^)   

 ^ ^  

 

    ^      

    ^  ^2  ^3 /^22 ( 4 0 )^3 /^2 3 ( 4 0 )^4 2 ( 4 4 ) ( )^1 3 ( 4 4 )^4 2  1    3

(^56) cu. units.

 

 

 

    

^ ^ 











     

     

      

   

  

  

2 0

2 3

2 0

2 2

(^222) 0

2 0

0 2

2 4 4 x x x dx

2 4 x x 4 x dx

2 y xy dx but x 4 y; y 4 - x

2 1 x ydx

V 2 1 x ydx

dV 2 ( 1 x) ydx

       

2 3 4 2

0 2 3 4

2 4 4 2 3 4 1 1 2 4 2 2 2 2 2 3 4 8 2 8 8 4 3

x x x

V  x

     (^)      (^)   

            ^ 

         ^ 

2 cu. units

 ^ 

^ ^ ^ 

HOMEWORK

A. Using disk or ring method, find the volume generated by revolving about the indicated axis the areas bounded by the following curves: 1.y = x3, y = 0, x = 2; about x-axis 2.y = 6x – x2, y = 0; about x-axis 3.y2 = 4x, x = 4; about x = 4 4.y = x2, y2 = x; about x = - 5.y = x2 – x, y = 3 – x2; about y = 4

B. Using cylindrical shell method, find the volume generated by revolving about the indicated axis the areas bounded by the following curves:

1.y = 3x – x2, the y-axis, y = 2; about y-axis

3. y = x3, x = y3; about x-axis

, 8

1

2. (^) y  4 x  x^4 y-axis, about x=