Use - disks, washers or shells? integrate volume of revolving solids

[Rattanjeet]

 

[Ackbach]

Disks and washers are essentially equivalent - it's analogous to finding area versus finding the area between two curves. So just thinking disks verses washers: it depends on the shape of the relation (not necessarily functions!), what is the best method to use. If you have a relation that's also a function and even 1-1, then it probably doesn't matter. If you have a function that's not 1-1, try [itex]dx[/itex] disks. If you have [itex]x[/itex] as a function of [itex]y[/itex], but it's not 1-1, try shells with a [itex]dy[/itex]. You don't want the infinitesimal shape to intersect the relation except at the endpoints, or you'll have to break up your integral into multiple pieces, which ends up being more work.

 

[Rattanjeet]

How do I set up the integral for the curve y = x, x = 2 - y^2, and y = 0 to find the volume
when it is revolved

around x axis
around y axis
about the line x = -1
y = 1

I am quite confused. Do I change x = 2 - y^2 to y = Squareroot (2 - x), when required. If I do so, does the limit of integral also change or does it remain the same.

 

[Ackbach]

You will need to find the intersections of all three defining curves. Where does that happen?

For the around the x-axis case, I'd recommend shells using a dy.

For the around the y-axis case, I'd recommend washers using a dy. Use the same for about the line x = -1.

For around the line y=1, I'd recommend shells using a dy.

Can you see why I picked these?

 

[Mark44]

How do I set up the integral for the curve y = x, x = 2 - y^2, and y = 0 to find the volume
when it is revolved

around x axis
around y axis
about the line x = -1
y = 1

I am quite confused. Do I change x = 2 - y^2 to y = Squareroot (2 - x), when required. If I do so, does the limit of integral also change or does it remain the same.

The following is copied from a PM sent by Rattanjeet.

I think I cannot make the graph to show the region bounded by X = 2-y^2, Y=0, and y=x

on setting y = 2 - y^2, I get the limits of integration ±y.

Given below is my work; this is how I set up the integrals:

a) Volume when the region is rotated about x-axis:

Method used: shell method

V=2∏∫[c-d] y g(y) dy
V=2∏{∫[0-1] y (y) dy - ∫[1-2] y(2-y^2)} dy
V=2∏{∫[0-1] y^2 dy - ∫[1-2] (2y-y^3)} dy

b) Volume when the region is rotated about y-axis:

Method Used: Washer Method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V= ∏{∫[0-1] (2-y^2)^2dy - ∫[1-2] (y)^2 dy }

c) Volume when the region is rotated about x = -2:

Method used: Wahser method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V= ∏{∫[0-1] (2-y^2 + 2)^2dy - ∫[1-2] (y+2)^2 dy }
V= ∏{∫[0-1] (4-y^2)^2dy - ∫[1-2] (y+2)^2 dy }

d) Volume when the region is rotated about y = 1:

Method Used: Washer Method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V = ∏∫[0-1] (1 - y)^2dy - ∏∫[0-1](1-(2-y^2))^2 dy
V = ∏∫[0-1] (1 - y)^2dy - ∏∫[0-1](y^2-1))^2 dy

I just want guidance as to how to determine which method to use and when; also when to change the function from y= f(x) to x = f(y) and vice versa.

I have read the book again and again, solved dozens of questions and still I am not sure. I want to be confident when I apply the method and set up the integral and its limits. This is the guidance I require. I do not want any question to be solved for me.

 

[Rattanjeet]

Thanks for your suggestion.

Intersections are (0,0), (1,1) and (2,0)

And my working of this question is as follows:

a) Volume when the region is rotated about x-axis:

Method used: shell method

V=2∏∫[c-d] y g(y) dy
V=2∏{∫[0-1] y (y) dy - ∫[1-2] y(2-y^2)} dy
V=2∏{∫[0-1] y^2 dy - ∫[1-2] (2y-y^3)} dy

b) Volume when the region is rotated about y-axis:

Method Used: Washer Method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V= ∏{∫[0-1] (2-y^2)^2dy - ∫[1-2] (y)^2 dy }

c) Volume when the region is rotated about x = -2:

Method used: Wahser method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V= ∏{∫[0-1] (2-y^2 + 2)^2dy - ∫[1-2] (y+2)^2 dy }
V= ∏{∫[0-1] (4-y^2)^2dy - ∫[1-2] (y+2)^2 dy }

d) Volume when the region is rotated about y = 1:

Method Used: Washer Method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V = ∏∫[0-1] (1 - y)^2dy - ∏∫[0-1](1-(2-y^2))^2 dy
V = ∏∫[0-1] (1 - y)^2dy - ∏∫[0-1](y^2-1))^2 dy

Is my approach correct?

Rattanjeet

You will need to find the intersections of all three defining curves. Where does that happen?

For the around the x-axis case, I'd recommend shells using a dy.

For the around the y-axis case, I'd recommend washers using a dy. Use the same for about the line x = -1.

For around the line y=1, I'd recommend shells using a dy.

Can you see why I picked these?

 

[Mark44]

Let's take a look at the first two parts of your question. Generally it is better to ask a single question in a thread.

Intersections are (0,0), (1,1) and (2,0)

These are correct.

And my working of this question is as follows:

a) Volume when the region is rotated about x-axis:

Method used: shell method

V=2∏∫[c-d] y g(y) dy
V=2∏{∫[0-1] y (y) dy - ∫[1-2] y(2-y^2)} dy
V=2∏{∫[0-1] y^2 dy - ∫[1-2] (2y-y^3)} dy

Your setup in the 2nd line above is incorrect, and will give you the negative of the volume. For the length of the shell, you want x_parabola - x_line. You have it the other way around, which produces a negative value.

In the shell method you are taking thin horizontal strips that extend from the line x = y to the parabola x = 2 - y². The length of a strip is 2 - y² - y. The width of each strip is Δy, a small subinterval in the interval [0, 1] on the y-axis.

ΔV, the volume of a typical volume element, is given by ΔV ≈ 2##\pi## [radius] * [length of strip] * Δy

This problem could also be done using washers, but you would need two integrals.

b) Volume when the region is rotated about y-axis:

Method Used: Washer Method

V = ∏∫[c-d] f(y)^2 - g(y)^2 dy
V= ∏{∫[0-1] (2-y^2)^2dy - ∫[1-2] (y)^2 dy }

This is wrong as well. The typical volume element here is
ΔV ≈ ##\pi## [x_outer² - x_inner²] * Δy
= ##\pi## [(2 - y²)² - y²]Δy

You don't need two integrals.

 

[Rattanjeet]

Thanks Mark,

But I have to ask you how to remove a thread that I posted just today. To this very question I got another reply saying that my answer was correct and I should put it on thread. I did that. Since the answer was wrong, I would like to withdraw it so that others who read it are not misled.

Please guide.

Thanks once again.
Rattanjeet

 

[Mark44]

You have a total of 4 posts, and they're all in this thread. I deleted one of your posts, as it seemed to be a duplicate of what is already in this thread.

 

[Mark44]

I should also mention that it's really important to draw a sketch of the region you're working with, and another sketch of the solid you get by revolving the region. It is very difficult to get these problems right if you don't have a reasonably good picture.