Question 1015808
<pre>
The other tutor has set up the integrals for you. 
I'll draw some graphs and pictures to help you.

The whole graph of xy=1 looks like this

{{{graph(400,144,-6,4,-1.8,1.8,y=1/x)}}}

But we only use the one little sliver of it between x=1 and x-2.

This is the region (area) to rotate, and you are to rotate it about the
green line x=-1:
 
{{{drawing(400,144,-6,4,-1.8,1.8, 
red(line(1,0,1,1),line(2,0,2,.5)),
green(line(-1,-3,-1,3)),
 
graph(400,144,-6,4,-1.8,1.8,
 


(1/x)*(sqrt(x-1)/sqrt(x-1))*(sqrt(2-x)/sqrt(2-x)))




 
)}}}
 
When it gets rotated, it will look like this:
 
{{{drawing(400,144,-6,4,-1.8,1.8, 
red(line(2,0,2,.5)),red(line(-4,0,-4,.5)),
green(line(-1,-3,-1,3)),
 
red(arc(-1,1,4,-.5)),red(arc(-1,0,6,-.75,180,360)),
 locate(-2,1.2,HOLE),
red(arc(-1,.5,6,-.75,180,360)),
 
graph(400,144,-6,4,-1.8,1.8,
 


(1/x)*(sqrt(x-1)/sqrt(x-1))*(sqrt(2-x)/sqrt(2-x))),
 
graph(400,144,-6,4,-1.8,1.8,
 


(-1/(x+2))*(sqrt(x+4)/sqrt(x+4))*(sqrt(-3-x)/sqrt(-3-x))




 
))}}}
 
It will have a big hole cut in the center.  You are instructed 
to use the washer method.  That's why the other tutor used two
integrals.  The figure has to be  broken into two parts.  If we 
could use the cylindrical shell method, we would not have to do 
that.  But since your instructions specify that you are to use
the washer or disk method, then we have no choice but to break it 
into two parts. Here are the two parts that represent the two
integrals the other tutor used.:


{{{drawing(400,144,-6,4,-1.8,1.8, 


green(line(-1,-3,-1,3)),
 
red(arc(-1,1,4,-.5)),
 locate(-2,1.2,HOLE),
red(arc(-1,.5,6,-.75,180,360)),
 
graph(400,144,-6,4,-1.8,1.8,
 


(1/x)*(sqrt(x-1)/sqrt(x-1))*(sqrt(2-x)/sqrt(2-x))),
 
graph(400,144,-6,4,-1.8,1.8,
 


(-1/(x+2))*(sqrt(x+4)/sqrt(x+4))*(sqrt(-3-x)/sqrt(-3-x))




 
))}}}{{{drawing(400,144,-6,4,-1.8,1.8, 
red(line(2,0,2,.5)),red(line(-4,0,-4,.5)),
green(line(-1,-3,-1,3)),
 
red(arc(-1,0,6,-.75,180,360)),


red(arc(-1,.5,6,-.75,0,360)),
 


 
graph(400,144,-6,4,-1.8,1.8)




 
))}}}


The reason we have to break it up is because the right end
of a typical washer for the upper part of the solid is on 
the graph of xy = 1, but a typical washer for the bottom
part of the solid is on the line x = 2.  If we could take our
element of area vertical instead of horizontal, as we would
if we could use the cylindrical shell method, the top of the
element of area would all lie on the curve xy = 1, and we 
would not have to break the solid.  I am surprised that
your teacher gave you this problem to do by the disk/washer
method instead of waiting until you had studied the cylindrical
shell method, and assigned it to be done that way. 
A typical washer for the upper part is


{{{drawing(400,144,-6,4,-1.8,1.8, 





green(line(-1,-3,-1,3)),
 
red(arc(-1,.75,4,-.6)),


red(arc(-1,.65,4,-.6,180,360)),




red(arc(-1,.75,5,-.75)),


red(arc(-1,.65,5,-.75,180,360)),
 
graph(400,144,-6,4,-1.8,1.8,
 


(1/x)*(sqrt(x-1)/sqrt(x-1))*(sqrt(2-x)/sqrt(2-x))),
 
graph(400,144,-6,4,-1.8,1.8,
 


(-1/(x+2))*(sqrt(x+4)/sqrt(x+4))*(sqrt(-3-x)/sqrt(-3-x))




 
))}}}

A typical washer for the bottom part:

{{{drawing(400,144,-6,4,-1.8,1.8, 


red(line(2,0,2,.5)),red(line(-4,0,-4,.5)),


green(line(-1,-3,-1,3)),
 

red(arc(-1,.35,4,-.6)),


red(arc(-1,.25,4,-.6,0,360)),



red(arc(-1,.35,6,-.75)),


red(arc(-1,.25,6,-.75,0,360)),
 
graph(400,144,-6,4,-1.8,1.8)
 




 
))}}}




 


Edwin</pre>