# Volume of Revolution

1. Jan 16, 2004

### Little Dump

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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2. Jan 17, 2004

### himanshu121

It is so small to understand the pic u quoted so i'm giving u my solution

3. Jan 17, 2004

### himanshu121

I hope u will take it from here

http://in.geocities.com/mathsforjee/index.htm [Broken]

Last edited by a moderator: May 1, 2017
4. Jan 17, 2004

### Little Dump

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

I'll keep trying to figure it out.

5. Jan 17, 2004

### HallsofIvy

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.

6. Jan 17, 2004

### himanshu121

Is this true for all kind of figure I never came across that theorem is there any link where i can go for reference

7. Jan 18, 2004

### HallsofIvy

Pappus' theorem is true of any "solid of revolution". You should be able to find it in any calculus textbook (that includes multiple integrals.)

8. Jan 18, 2004

### 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.

$$\int \pi r_o^2 - \pi r_i^2 dr$$

where

$$r_o=(-y+3)$$
$$r_i=(y+1)$$

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

so we have this

$$\int_0^1 \pi (-y+3)^2 - \pi (y+1)^2 dy$$

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.

Last edited: Jan 18, 2004
9. Jan 18, 2004

### himanshu121

Here is the general formula

http://in.geocities.com/mathsforjee/GM.html [Broken]

Last edited by a moderator: May 1, 2017
10. Jan 18, 2004

### 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

11. Jan 18, 2004

### himanshu121

Yes it do makes sense dont you have gone to the previous post.

Thats what u have to do and its general too

12. Jan 18, 2004

### himanshu121

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

13. Jan 18, 2004

### Little Dump

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?

14. Jan 18, 2004

### Little Dump

I'm rotating horizontal rectangles around the y-axis therefore each rectangle will make a circle

15. Jan 18, 2004

### himanshu121

Ok It would form rings as of saturn

16. Jan 18, 2004

### himanshu121

U can also do it analytically With no integration

17. Jan 18, 2004

### Little Dump

I still want to know whats wrong with this answer because it does not yield 10pi/3

$$\int_0^1 \pi (-y+3)^2 - \pi (y+1)^2 dy$$