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:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{a}^{b} f(x) dx[/tex] is the area under the graph.

[tex]\frac {\int_{a}^{b} f(x) dx} {b-a}[/tex] 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 [tex]\pi r^{2} (b-a)[/tex], where [tex]r[/tex] is [tex]\frac {\int_{a}^{b} f(x) dx} {b-a}[/tex].

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 [tex]V = \frac {\pi (\int_{a}^{b} f(x) dx)^2} {b-a}[/tex]?

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

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# Homework Help: Volume of rotated graph

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