Expert: Paul Klarreich - 11/8/2006

Question
Find the area of the surface generated when y=((x^3)/12)+(1/x) from x=1 to x=2 is rotated about the x-axis.  Instructor wasn't totally clear on formula of surface of revolution.  If you could help out I appreciate it.  Thank you.

Answer
Questioner:  Gary
Category:  Advanced Math
 
Subject:  Surface of a revolution

>> You mean 'A surface of revolution.'  You want to know about 'a revolution'?  Take a history class.

Question:  Find the area of the surface generated when

y= x^3/12 + 1/x from x=1 to x=2 is rotated about the x-axis.  Instructor wasn't totally clear on formula of surface of revolution.  

>> He's probably young.

If you could help out I appreciate it.  

Thank you.
.........................................
Hi, Gary,

To do any application of integration, you look at a sample piece.  For the surface of revolution, take an itty-bitty piece of the graph of f(x) on [a,b] and rotate it around the x-axis.  That gives you a small ring, and the surface area of this 'ring' is given by:

2 pi r ds, where:

ds = the element of arc length and = sqrt(1 + f'(x)^2) dx
r = the average radius, which we can take as  f(x).

So your surface is:

{b
| 2 pi f(x) sqrt(1 + [f'(x)]^2) dx  << Standard SA formula.
}a

[USUAL WARNING ABOUT FIXED-SIZE FONTS.]

In your example, you have f(x) = x^3/12 + 1/x,  a = 1,  b = 2.  You also need:
                       x^4 -  4
f'(x) = x^2/4 - 1/x^2 = --------
                         4x^2

Looks messy, but these things have a way of working themselves out.  If they didn't, textbook authors would become murder victims -- the students would kill them.

           (x^4 - 4)^2
[f'(x)]^2 = -------------
              16x^4

           x^8 - 8x^4 + 16
[f'(x)]^2 = ---------------
                16x^4

               x^8 - 8x^4 + 16
1 + [f'(x)]^2 = --------------- + 1
                    16x^4

               x^8 - 8x^4 + 16 + 16x^4
1 + [f'(x)]^2 = -----------------------
                    16x^4


               x^8 + 8x^4 + 16
1 + [f'(x)]^2 = ---------------
                    16x^4

               (x^4 + 4)^2
1 + [f'(x)]^2 = ------------
                  16x^4
            x^4 + 4
sqrt(that) = -------
              4x^2

OK, preliminary work done, proceed to surface area:

{b
| 2 pi f(x) sqrt(1 + [f'(x)]^2) dx      << Again.
}a


{2                    x^4 + 4  
| 2 pi [x^3/12 + 1/x] -------  dx
}1                     4x^2


{2     x^4 + 12   x^4 + 4  
| 2 pi ---------- -------  dx
}1        12x       4x^2


{2     x^4 + 12   x^4 + 4  
|   pi ---------- -------  dx
}1         6x       4x^2


{2  pi  (x^4 + 12)(x^4 + 4)  
|  ---  -------------------  dx
}1  24        x^3


{2  pi  x^8 + 16x^4 + 48  
|  ---  ----------------  dx
}1  24        x^3


{2  pi  
|  --- [x^5 + 16x + 48x^-3] dx
}1  24        

Whew!  Now we can integrate:
pi  x^6          48x^-2
--- [--- + 8x^2 + ------ ]
24   6            -2


pi  x^6            24
--- [--- + 8x^2 - ------ ]  at  x=2 and x=1
24   6            x^2

pi
---- [ [64/6 + 32 - 24/4]  -  [1/6 + 8 - 24] ]
24

pi
---- [ 32/3 + 26  -  1/6 + 16 ]
24


pi    64 + 156 - 1 + 96
---- [ ----------------- ]
24              6

pi    315
---- [ ---]
24     6

pi    105
---- [ ---]
24     2
 105 pi
= -------
   48

I looked at the graph of your f(x).  From x=1 to x=2 it stays pretty close to y=1.  If you take that segment, about 1 unit long, with an average height about 1 unit, and rotate it, the resulting 'cylinder' has surface area a little more than 2pi.  That's just about what we have.