Calculating Volume of Solid Using Shells Method - Casey

[Saladsamurai]

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

 

[Saladsamurai]

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

 

[bob1182006]

yes that's right.

so your integral will be:

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 ><

 

[Saladsamurai]

yes that's right.

so your integral will be:

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 ><

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

 

[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).

 

[Saladsamurai]

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).

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.

 

[bob1182006]

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

 

[Saladsamurai]

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!

 

[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.

 

[Saladsamurai]

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.

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 wherever there was an x..like this:
2\pi\int_0^1(x+2)[(2-(x+2))^2-(x+1)^2]dx

 

[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 ?

 

[Saladsamurai]

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

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

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. I guess that's what you meant by the washer ?

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?

 

[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

 

[Saladsamurai]

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.

 

[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.

 

[Saladsamurai]

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.

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

 

[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.

 

[Saladsamurai]

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.

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

Thanks fo ryour help Bob,
Casey

 

[bob1182006]

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

 

[Saladsamurai]

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

 

[bob1182006]

hm..how did you get 3-sqrty? same with 3-(2-y).

and the limits on both integrals should be the same.
and the terms inside the integral should be squared since you're not using 2pi xf(x)

 

[Saladsamurai]

hm..how did you get 3-sqrty? same with 3-(2-y).

and the limits on both integrals should be the same.
and the terms inside the integral should be squared since you're not using 2pi xf(x)

I don't know..I wrote it w.r.t. y...what is happening to me?

I am going to upload a picture...

Casey

 

[bob1182006]

duno should be:
x=sqrt(y)
x=2-y

limits:
x=0
x=1

and since you're doing it around x=3. you should substitute x+3 into all equations that have x in it.

and then plug into the integral.

 

[Saladsamurai]

duno should be:
x=sqrt(y)
x=2-y

limits:
x=0
x=1

and since you're doing it around x=3. you should substitute x+3 into all equations that have x in it.

and then plug into the integral.

We have never written an integral in terms of y with integrtion limits in terms of x??

Here is the picture.

R_2 is just from the y-axis to the line x=3...R_1 is from (x) to the line x=3...but if height is \Delta y the values of x fall on different curves namely from y=0-->1, x falls on sqrt(y) and from y=1-->2, x falls on 2-y...

Thus, from y=0-->1, R_1 =3-(sqrt(y)) and from y=1-->2 R_1=3-(2-y)

I don't see why I would not split the integral?

Casey

 

[bob1182006]

Hm..yea I see what you're doing, but the integral you wrote would just be the area enclosed by those formulas * pi.

so you should just square the terms inside the integral in order to give you the area of revolution.

I just never did areas of revolution this way :s

 

[Saladsamurai]

Hm..yea I see what you're doing, but the integral you wrote would just be the area enclosed by those formulas * pi.

so you should just square the terms inside the integral in order to give you the area of revolution.

I just never did areas of revolution this way :s

Yes..I meant to write
V=\pi\int_0^1(3-\sqrt{y})^2dy+\pi\int_1^2[3-(2-y)]^2dy <-----this is like finding the volume of the big cylinder whose radius extends from x=0->3 and removing the volume of the empty space inside of it.

I am sorry if that made this more difficult than it had to be...I just don't see another way to do this (less the shells method) that I have learned. I have never seen x's and y's mixed together in the same integral before.

Casey

 

[bob1182006]

kay yea I see it now o.o.

The other way (shells?) would be like:
y=x^2
y=2-x

limits:
x=0
x=1

about x=3 so substitute x+3 into everything, so now you're rotating about the y-axis.

y=(x+3)^2=x^2+6x+9
y=2-(x+3)=2-x-3=-(x+1)=-x-1

limits:
x+3=0, x=-3
x+3=1, x=-2

-(x+1) is still above (x+3)^2

so the integral is:
2\pi \int_{-3}^{-2}x((-x-1)-(x^2+6x+9)) dx = 2\pi \int_{-3}^{-2} (-x^2-x-x^3-6x^2-9x) dx

which seems simpler imo but maybe the other way is more intuitive? either way should be the same answer

 

[Saladsamurai]

kay yea I see it now o.o.

The other way (shells?) would be like:
y=x^2
y=2-x

limits:
x=0
x=1

about x=3 so substitute x+3 into everything, so now you're rotating about the y-axis.

y=(x+3)^2=x^2+6x+9
y=2-(x+3)=2-x-3=-(x+1)=-x-1

limits:
x+3=0, x=-3
x+3=1, x=-2

-(x+1) is still above (x+3)^2

so the integral is:
2\pi \int_{-3}^{-2}x((-x-1)-(x^2+6x+9)) dx = 2\pi \int_{-3}^{-2} (-x^2-x-x^3-6x^2-9x) dx

which seems simpler imo but maybe the other way is more intuitive? either way should be the same answer

So when you do it this way ^^^ are you just "shifting" everything back toward the y-axis? Hmmm...

Anyway thanks again Bob. I will reread this thread agin in the a.m. I need sleep.

thanks again,
Casey

 

[bob1182006]

no problem.

and yes if you have it rotated about x=3, you can just do a translation to the function to get it rotated about the y or x-axis which simplifies things.

Of course you have to shift the limits and the functions also. But imo it's easier than drawing a picture and then finding the radius :s but w/e works for ya!