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Volume of Revolution

by Little Dump
Tags: revolution, volume
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Little Dump
#1
Jan16-04, 11:30 PM
P: 19
Included is my attempt at the following question. I get an answer of 10pi, whereas the right answer is (10pi)/3 from my text book. Here is the question:

Rotate the triangle described by (-1,0),(0,1),(1,0) around the axis x=2 and calculate the volume of the solid.

I basically changed the problem to the following and continued as it is the same

Rotate the triangle described by (1,0),(2,1),(3,0) around the y-axis and calculate the volume of the solid

Thanks for the help
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himanshu121
#2
Jan17-04, 12:10 AM
himanshu121's Avatar
P: 658
It is so small to understand the pic u quoted so i'm giving u my solution
himanshu121
#3
Jan17-04, 01:31 AM
himanshu121's Avatar
P: 658
I hope u will take it from here

http://in.geocities.com/mathsforjee/index.htm

Little Dump
#4
Jan17-04, 08:18 AM
P: 19
Volume of Revolution

I don't quite understand it and I dont understand why mines wrong :(

I'll keep trying to figure it out.
HallsofIvy
#5
Jan17-04, 08:26 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682
You may not have had it yet but this looks like an exercise in using Pappus' theorem: the volume of a solid of revolution is the area of a cross section times the circumference of the circle generated by the centroid of that cross section.

In this case, the cross section is a triangle with base of length 2 and height 1: area= (1/2)(2)(1)= 1.
The centroid (for a triangle only) is the "average" of the vertices:
((-1+0+1)//3,(0+1+0)/3)= (0, 1/3). The distance from (0, 1/3) to the line x= 2 is 2- 1/3= 5/3. The centroid "travels in" (generates) a circle of radius 5/3 and so circumference (10/3)pi.

The volume of the figure is (1)(10/3)pi= (10/3)pi.
himanshu121
#6
Jan17-04, 11:53 PM
himanshu121's Avatar
P: 658
You may not have had it yet but this looks like an exercise in using Pappus' theorem: the volume of a solid of revolution is the area of a cross section times the circumference of the circle generated by the centroid of that cross section.
Is this true for all kind of figure I never came across that theorem is there any link where i can go for reference
HallsofIvy
#7
Jan18-04, 06:07 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682
Pappus' theorem is true of any "solid of revolution". You should be able to find it in any calculus textbook (that includes multiple integrals.)
Little Dump
#8
Jan18-04, 08:32 AM
P: 19
I haven't learned that theorem so I don't really think I should use it.

I'm trying to use horizontal rectangles for my area so I formulated the following integral which is how I was taught how to do questions like this.

[tex]
\int \pi r_o^2 - \pi r_i^2 dr
[/tex]

where

[tex]r_o=(-y+3)[/tex]
[tex]r_i=(y+1)[/tex]

and the limits of integration are from y=0 to y=1

so we have this

[tex]
\int_0^1 \pi (-y+3)^2 - \pi (y+1)^2 dy
[/tex]

it makes perfect sense to me but then once you work it out you get 10 pi. which is wrong

So can someone point out what I did wrong and how to fix it so I dont do it again.


Thanks very much.
himanshu121
#9
Jan18-04, 08:39 AM
himanshu121's Avatar
P: 658
Originally posted by Little Dump
I haven't learned that theorem so I don't really think I should use it.

I'm trying to use horizontal rectangles for my area so I formulated the following integral which is how I was taught how to do questions like this.

[tex]
\int_0^1 \pi r_o^2 - \pi r_i^2
[/tex]

fjsjf
Here is the general formula

http://in.geocities.com/mathsforjee/GM.html
Little Dump
#10
Jan18-04, 08:43 AM
P: 19
does mine not make sense for some reason

i am calculating the area of the circle when the farthest line is rotated and then subtracting the area of the circle when the closest line is rotated
himanshu121
#11
Jan18-04, 08:49 AM
himanshu121's Avatar
P: 658
Yes it do makes sense dont you have gone to the previous post.

Thats what u have to do and its general too

Your way do make sense
himanshu121
#12
Jan18-04, 08:52 AM
himanshu121's Avatar
P: 658
Originally posted by Little Dump
does mine not make sense for some reason

i am calculating the area of the circle when the farthest line is rotated and then subtracting the area of the circle when the closest line is rotated

When you rotate a point about a line u get a circle

Not

When you rotate a line

It would be somewhat like a truncated cone
Little Dump
#13
Jan18-04, 08:53 AM
P: 19
but i get the wrong answer so can you point out whats wrong with my formulation?

keep in mind i moved the triangle to (1,0),(2,1),(3,0) because its the same problem correct?
Little Dump
#14
Jan18-04, 08:56 AM
P: 19
When you rotate a point about a line u get a circle

Not

When you rotate a line

It would be somewhat like a truncated cone

I'm rotating horizontal rectangles around the y-axis therefore each rectangle will make a circle
himanshu121
#15
Jan18-04, 09:01 AM
himanshu121's Avatar
P: 658
Originally posted by Little Dump
I'm rotating horizontal rectangles around the y-axis therefore each rectangle will make a circle
Ok It would form rings as of saturn
himanshu121
#16
Jan18-04, 09:05 AM
himanshu121's Avatar
P: 658
U can also do it analytically With no integration
Little Dump
#17
Jan18-04, 09:33 AM
P: 19
I still want to know whats wrong with this answer because it does not yield 10pi/3

[tex]\int_0^1 \pi (-y+3)^2 - \pi (y+1)^2 dy[/tex]


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