What was the calculus formula to determine the volume of a solid? I don't mean solid of revolution. Just regular solid.

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What was the calculus formula to determine the volume of a solid? I don't mean solid of revolution. Just regular solid.

Mathematics
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volumes of cross sections
Are you sure you aren't looking for: \[\text{ Volume }=\int\limits_a^bA(x) dx \text{ where } A(x) \text{ is the area of the cross section } ?\]
yes that is it ! thank you !

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that's also the formula used in solids formed by revolution of some curve about some axis
u forget to square it..i think @freckles
you know where we take circles as the cross section
no A(x) is the area you don't want to square the area
o ok
if the cross section is a circle then A(x)=pi*r^2
where r is a function of x
though we could look at things in terms of y too not just x
you know depending on what suits us better
but how is that the formula for solids of revolution? I mean the formula for solids of revolution is \[\int\limits_{a}^{b}(\pi*f(x)^2) dx\]
i think the f(x) is actually the radius
f(x) is the function
i mean the height of the funtion
yes which is intern the radius so A=pi*r^2
and freckles just use the Area instead...i think
I guess i am kind of confuse lol
ok lets wait on till @freckles come online...
what's the question?
area of a circle is what I called A(x)
or the are of the cross section is A(x)
if the cross section is a circle then A(x) is pi*r^2 where r is the radius f(x) is the distance from the curve to the x-axis so f(x) is the radius if the axis of rotation is the x-axis
|dw:1440894543729:dw| Say we want to take this curve on [a,b] and rotate it about the x-axis
first reflecting the curve about the x-axis gives us: |dw:1440894612771:dw| now that drawing is not that great but that curve is basically suppose to be a mirror for the other curve I drew
|dw:1440894662165:dw| so we have a cross section that is a circle
notice the distance from the x-axis to the curve is indeed the radius which is called f(x) in this case
\[\int\limits_{a}^{b}A(x) dx \\ \text{ so we have } A(x)=\pi \cdot (f(x))^2 \\ \]
so i was right about f(x) being the radius?
ohh i see
|dw:1440894830121:dw| yes
but the A(x) is not squared
A(x) just means area of the cross section
ohh i see ! I thought f(x) was just the height of the function
It makes so much sense now :)
well you can also say |f(x)| is the height of the function
I put | | around it because f(x) could happen to be negative which just means the height is below the x-axis
cool..

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