# 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

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

Staff Emeritus
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

Staff Emeritus
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

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

Your way do make sense

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

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