+0

# ​ Volume of revolution

0
308
2 So far in my attempt i have found the normal area of the shaded region then realising i cannot convert it into the volume of revolution (or i don't know how to)

Jan 23, 2019

#1
+2

We can use the "Ring Method" to solve this

The volume is   =  volume of top function rotated about the x axis from x = .5 to x = 1   less the  volume of the bottom function rotated about the x axis  from x = .5 to x = 1

So  we have

1                              1

pi  ∫   (x + 1/x)^2 dx    -    pi  ∫  ( 2x)^2 dx    =

.5                            .5

1                                          1

pi  ∫   x^2 + 2 + 1 /x^2  dx  -  pi  ∫  4x^2   dx    =

.5                                         .5

1

pi  ∫   2  + 1/x^2  - 3x^2   dx  =

5

1              1             1

pi (    2x ]      - 1 /x ]     -  x^3 ]    )   =

.5              .5            .5

pi * ( [ 2 - 1 ] - [ 1 - 2 ] - [ 1 - .125 ]  )  =

pi  ( 1 +  1  - 0.875  )  =

[  1.125 pi   ]  units^3   Jan 23, 2019
edited by CPhill  Jan 23, 2019
#2
+1

noted!!

thank you very much

YEEEEEET  Jan 23, 2019