Question 594147
find the volume bounded by the x-axis, y=x^2+1, and x=-1,x=1 revolved about the x-axis.
<pre>
Here is the area to be revolved.  When revolved about the x-axis 
a pulley will be created with no hole for an axle.

{{{drawing(300,500,-1.5,1.5,-2.5,2.5,

rectangle(.6,0,.65,1.36),

graph(300,500,-1.5,1.5,-2.5,2.5,(x^2+1)*(sqrt(1-x)/sqrt(1-x))*

(sqrt(x+1)/sqrt(x+1))), red(line(1,0,1,2),line(-1,0,-1,2)),
locate(.6,0,dx), locate(.655,.7,y)

 )}}}

The vertical elements all have one end on the same graph (the x-axis)
and the other end on the same graph (the parabola).  We could not say the
same if we used horizontal elements.  

We can do it this way:

Volume = {{{pi*int((radius^2*thickness),"",lower,upper) }}} = {{{pi*int(y^2,dx,-1,1) }}} = {{{pi*int((x^2+1)^2,dx,-1,1) }}} = {{{pi*int((x^4+2x^2+1),dx,-1,1) }}} =

But because of symmetry w/r the y-axis we can just find the volume 
of the right side of the pulley and double it, and take the lower limit 
as 0 instead of -1.

{{{2pi*int((x^4+2x^2+1),dx,0,1) }}} =

{{{2pi*(int(x^4,dx)+2int(x^2,dx)+int(1,dx))}}}{{{matrix(3,2,"|",1,"|","","|",0)}}} =

{{{2pi*(x^5/5 +2x^3/3 + x)}}}{{{matrix(3,2,"|",1,"|","","|",0)}}} =

{{{2pi*(1^5/5 +2*1^3/3 + 1)}}} - {{{2pi*(0^5/5 +2*0^3/3 + 0)}}} =

{{{2pi*(1/5 +2/3 + 1)}}}

{{{2pi*(3/15 +10/15 + 15/15)}}} =

{{{2pi*(28/15)}}} =

{{{expr(56/15)pi}}}

Edwin</pre>