Expert: Paul Klarreich - 5/6/2007

Question
Rotate around the y-axis the region above the graph of y=x^3 that is bounded by the lines x=1 and y=8.

So, I was using V=2pixydx and put in x^3-7 for y. So, I am trying to integrate 2pi(x)(x^3-7)dx between x=1 and x=2. I am getting close to the answer, but not quite there, so not sure where I am off. Thanks for your help.

Answer
Questioner:   Colleen
Category:  Calculus
 
Subject:  Volume of a solid of revolution by cylindrical shells
Question:  Rotate around the y-axis the region above the graph of y=x^3 that is bounded by the lines x=1 and y=8.

So, I was using V= 2 pi x y dx and put in x^3-7 for y.

>> That part doesn't look right.  I think you want  8 - x^3  as the 'y', but it really isn't a 'y' but an 'h'.

So, I am trying to integrate 2pi(x)(x^3-7)dx between x=1 and x=2. I am getting close to the answer, but not quite there, so not sure where I am off. Thanks for your help.
.....................................................
Hi, Colleen,

You are looking at a 'sort of triangular' region:

 Left : x = 1
 Right: x = 2, but really the point (2,8)
 Top  : y = 8
 Bottom: y = x^3
 
and then rotating.  A shell has volume =  2pi rh dx.  

   r is the radius, which will be x.
   h is the height, which will be the upper function, y = 8, minus the lower function, y = x^3.

dV = 2 pi x (8 - x^3) dx

And your volume should be this integral:

{2
|   2 pi x (8 - x^3) dx
}1

{2
|   2 pi( 8x - x^4) dx
}1

2 pi(4x^2 - x^5/5)  from x = 1 to x = 2

2 pi[  (4(4) - 32/5) - (4 - 1/5) ]

2 pi[  16 - 32/5 - 4 + 1/5 ]

2 pi[  12 - 31/5 ]

2 pi[  60/5 - 31/5 ]

2 pi[  29/5 ]

= 58pi/5