What is the formula for calculating the volume of a rotated graph?

[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?

 

[quantumdude]

Hi. There are a few problems on my homework that involved the volume of a rotated solid.

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

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

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

 

[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 $4\pi /3$.

 

[KingNothing]

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

 

[HallsofIvy]

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