# Volume by Shells

1. Oct 14, 2007

I need to find the volume of the solid generated by rotating the region bounded by $$y=x^2$$ $$y=2-x$$ and the y axis around the y-axis by shells.

Does this sound correct? When I "open" a small shell up to reveal a small rectangle, the rectangle's height$$=dx$$, length$$=2\pi(x)$$ and width$$=y$$

So I will integrate wrt x and break the integral into two parts at x=1?

I am learning this from the text after missing a class, so any insight is appreciated,
Casey

2. Oct 14, 2007

:tongue2:

...maybe not...this looks all wrong! Ahhhhhhhh! I think it dhould be wrt y...maybe....

3. Oct 14, 2007

### bob1182006

yes that's right.

$$V=2\pi \int_a^b x(y_2-y_1) dx$$

and it's w.r.t. x because if you were to solve for x you'd have, x=+- sqrt y, and 2-y =x, which is 3 equations that enclose that volume.

Edit: whops wrong limits ><

Last edited: Oct 14, 2007
4. Oct 14, 2007

I do need to sum two integrals though correct i.e, where the top half's $$y_2-y_1=(2-x)-(1)$$ and the bottom half's $$y_2-y_1=1-(x^2)$$?

Casey

5. Oct 14, 2007

### bob1182006

no, y2 and y1 will be the functions that you are given. you just need to determine which one is the top(y2) and which is the bottom (y1). and then do y2-y1.

and the integral from my notes is like this:

$$2\pi \int_a^b xy_2 dx - 2\pi \int_a^b xy_1 dx$$

which shows the relation better I guess. Just easier to remember x(y2-y1).

6. Oct 14, 2007

Oh wow...I don't know why I thought I needed to break it....I am so confused by the shell's method.....

I thought that these little "strips' were going from left to right? So isn't y_2 and y_1 going to be on the same curve at some point.

Maybe I need to take a break and come back to this.:grumpy:

7. Oct 14, 2007

### bob1182006

yes y2 and y1 will intercept at a and b. so you need to find those points and use them as your limits.

8. Oct 14, 2007

So should my final integral be:

$$2\pi\int_0^1x[(2-x)-x^2]dx$$

..which I will simplify...

I hope that is correct...I just got a picture of it in my head and it made sense...

Say the y-axis is vertical wire pulled taught...then the "shells" would be like pieces of cloth wrapping around it. But the heights of the shells get narrower as I approach the outer layers. So what began as tall "sheets" being wrapped around the wire, are getting progressively narrower (vertically narrower, that is) as they move rightward (well not just rightward but with symmetry to the wire).

That might make sense....

Casey

p.s. Does the integral look right? Cause if not, that was way too much time invested in that analogy!:rofl:

9. Oct 14, 2007

### bob1182006

yes that's right.

Also if you get some problem like this one but instead of about the y-axis say it was supposed to be around x=2.
you should make a substitution x=x+2 into EVERY formula with x so you'll have the shell around x+2=2 -> x=0 -> y-axis. which would make the problem easier.

10. Oct 14, 2007

So for the identical region but revolved around x=2..that produces a washer right? (not that that matters, I just want the visual)....Do you mean just plug x+2 whereever there was an x..like this:
$$2\pi\int_0^1(x+2)[(2-(x+2))^2-(x+1)^2]dx$$

11. Oct 14, 2007

### bob1182006

not into the integral just the formulas so you'd have:

y=2-(x+2)=-x
y=(x+2)^2=x^2+4

you do that substitution mainly so you don't have to find the radius, so it always just equals x.

and the limits would have to be given since the functions never meet there's no way to find a and b by solving -x=x^2+4. Since in Calc II you deal just with real numbers. I guess that's what you meant by the washer ?

Last edited: Oct 14, 2007
12. Oct 14, 2007

Yes. That is why it is a washer. My next task is actually to revolve that same region around x=3! So, I don't have to change the limits do I?

13. Oct 14, 2007

### bob1182006

whops sorry made a mistake ><
(x+2)^2=/=x^2+4 >< so you can still do it and find the limits by making one equation equal to the other.

well you know the limits are:
x=0
x=1

and you will make a substitution x=x+p
so your new limits will be:
x+p=0
x+p=1

14. Oct 14, 2007

You know I just noticed that it does not say that this one needs to be done by shells.....just to find the volume.

I will probably opt to do it by $$\pi\int_a^b([f(x)]^2-[g(x)]^2)dx$$ then maybe I will attempt to use shells. I don't know if can be done by shells with the techniques I have learned so far...

Maybe if you have time you could check out what I get using the washer method. I'd rather know that it is correct, so when I go to do it by shells, I will have a comparitive answer.

I'll post my integral in a moment.

Last edited: Oct 14, 2007
15. Oct 14, 2007

### bob1182006

you shouldn't use that formula because it should be f(y)^2-g(y)^2.

you would use that integral when you have formulas like:
x=y^2
x=3-y

since that only has 2 formulas for x.

if you solve the current formulas for x you'll have 3 which won't work or would be harder.

16. Oct 14, 2007

I have used that many times in this course and not had to use$$^=_-\sqrt{y}$$

and it because we are only dealing with the region bounded by all three curves $$y=X^2$$ $$y=2-x$$ and $$x=0$$ i.e., the region in the first quadrant only.

This eliminates the -sqrt function, does it not?

Casey

17. Oct 14, 2007

### bob1182006

yes, but on some problems you might not be able to get rid of that -sqrt and have 3 equations for x.

and when you do that's when you can't use it that formula to find the volume.

18. Oct 14, 2007

I am with you now. I think I recall my proffesor telling us to choose the simplest method for this assignment...so I am assuming this would be the one. But, if I am feeling adventurous I will try the shells method (I also have a book on light that I checked out of the library, so I want to read that to

Casey

19. Oct 14, 2007

### bob1182006

no prob glad I could help, gave me a chance to re-read my calc 2 notes too xD

20. Oct 14, 2007

I cannot stand my brain right now...it is absolutely useless to me.

Is the integral for revolution about x=3:

$$V=\pi\int_0^1(3-\sqrt{y})dy+\pi\int_1^2[3-(2-y)]dy$$ ???

I am dying here ...ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh...

ah.

Casey