Thread: volume by shell method

This is a question on one of my past exams that I'm trying to do...

the question says to find the volume of the regions bounded by y=x^2-2 and
y=x

rotated about x=2

I was thinking that the radius equalls (2-x) [I'm not entirely sure about this]
and the height equals (x-(x^2-2))

all of them multiplied together with PIE under the integral from -1 to 2

is this correct?

Shells:

It's PI, not PIE for goodness sake.

why is it (x-2) and not (2-x)
finding the radius using this method is really confusing

It's the same thing.

Originally Posted by akhayoon

why is it (x-2) and not (2-x)
finding the radius using this method is really confusing

to avoid confusion, remember that we always consider the distance from the y-axis to the bounded region to be x (that is when we are revolving about a vertical line of course). the line x = 2 is to the right of the region, so the distance from the y-axis some point in the region is x, and the distance from that point to the line x = 2 is (2 - x). the (2 - x) is the radius of our revolution, so that's what we want.

yeah but the thing that confused me here is that the bounded region lies in both sides of the y-axis and not just right or left of it. so how would you be able to calculate the distance of the bounded region??

Originally Posted by akhayoon

find the volume of the regions bounded by y=x^2-2 and
y=x, rotated about x=2

the region looks like this:

Use the shell method:

Originally Posted by curvature

Use the shell method:

the formula you are using is for the disk method. and we are rotating about the vertical line x = 2, nto y = 2

Originally Posted by Jhevon

the formula you are using is for the disk method. and we are rotating about the vertical line x = 2, nto y = 2

Oh I'am sorry to make a mistake. Then the problem is harder than I expected.

But I still think the volume is a difference of two volumes. What are they in terms of integrals?

Originally Posted by curvature

But I still think the volume is a difference of two volumes. What are they in terms of integrals?

galactus has the right integral. it is the difference of two, but you usually combine it into one. the disk method can be done, but you would have to change the function to be in terms of y

Here is a graph of the region showing the 'shells'. I had it animated, but had to resize the picture. Then it wouldn't animate.

Originally Posted by galactus

Here is a graph of the region showing the 'shells'. I had it animated, but had to resize the picture. Then it wouldn't animate.

Beautiful graph! Thanks.
Hope the forum will allow animations to be run someday.

Actually, it does allow it...to a point. I have posted them before. But, with this one, I had to resize the picture. When I did, it wouldn't animate. I used to post animated graphs, but it seems lately it won't handle them unless I resize, then it doesn't work.