# Volume of rotated graph

1. Mar 24, 2006

### KingNothing

Hi. There are a few problems on my homework that involved the volume of a rotated solid. I do not know how to do these, but I'm trying to devise a method. This is what I figure:

$$\int_{a}^{b} f(x) dx$$ is the area under the graph.

$$\frac {\int_{a}^{b} f(x) dx} {b-a}$$ is the average height.
This is where logic comes in:
I figure you can use the mean value (average height) as the radius of an "average cylinder" of the function. Therefore, the volume would be $$\pi r^{2} (b-a)$$, where $$r$$ is $$\frac {\int_{a}^{b} f(x) dx} {b-a}$$.

Is this correct? I will simplify once i confirm my Latex is correct.

Alright, with this, is it okay to conclude that the over volume is $$V = \frac {\pi (\int_{a}^{b} f(x) dx)^2} {b-a}$$?

On a sidenote, I make mistakes quite often with Latex, is there an offline generator to check my syntax with?

Last edited: Mar 24, 2006
2. Mar 24, 2006

### Tom Mattson

Staff Emeritus
Do you mean a volume of revolution? It looks like it.

Not that I know of, but you do have 24 hours to edit your posts.

3. Mar 24, 2006

### AKG

No. Consider a sphere of radius 1, centered at the origin, so r as you've defined it would be $\pi /4$, since the area under f will just be the area of half of a circle of radius 1, which is $\pi /2$, and then you divide by b-a = 1-(-1) = 2 to get r. So

$$\pi r^2(b-a) = \pi ^3/8$$

On the other hand, you know the volume is $4\pi /3$.

4. Mar 25, 2006

### KingNothing

Mind helping me figure out how you actually do this then?

5. Mar 25, 2006

### HallsofIvy

Staff Emeritus
If you used the distance from the axis of rotation to the centroid of the figure instead of the "average radius" then you would have Pappus' theorem.
Here's a reference on volumes of revolution:
http://www.maths.abdn.ac.uk/~igc/tch/ma1002/int/node22.html