Find the volume of the resulting solid if the region under the curve y= 1/(x^2+3x+2) from x=0 to x=1 is rotated about (a) the x-axis and (b) the y-axis.

(Not bounded by y=0)

Question

Find the volume of the resulting solid if the region under the curve y= 1/(x^2+3x+2) from x=0 to x=1 is rotated about (a) the x-axis and (b) the y-axis.

(Not bounded by y=0)

roadjester · Answer

$$1/(x^{2}+3x+2)$$

roadjester · Answer

There are no other restrictions.

turingtest · Answer

always nice to have a picture of the area we are looking at|dw:1333301688461:dw|

turingtest · Answer

...something like that ;)

turingtest · Answer

around the x axis I recommend disk method
are you familiar with it?

roadjester · Answer

It is not bounded by y=0...

roadjester · Answer

Yes, I am familiar with disk, washer, and cyndrilical shells.

turingtest · Answer

then it is infinite, hooray!

roadjester · Answer

But only to a certain extent.

roadjester · Answer

That's a good thing?

turingtest · Answer

how do you figure?
if it is not bounded below by y= something the area is infinite, which means any solid you can get from revolving around a line is infinite, so I'm going to say they mean the positive portion of y.

roadjester · Answer

Huh? So when revolved around the y axis, the answer is infinity?

turingtest · Answer

of course
if the area is infinite how could the volume obtained by spinning it around something be finite?

roadjester · Answer

Then what about part a: the x-axis?

roadjester · Answer

Mind you, this is an even question so I don't have the answers.

turingtest · Answer

if this is the exact wording of the question it is poor
it says the area "under the function"
but that could be either in the x direction or y  direction, so that is lame...
I'm pretty sure they mean this region|dw:1333302430100:dw|