14 Jun 10
Originally posted by mlpriorI think I found a solution:
Anyone remember how to integrate to find surface area. Ugh.
For example, a 150 ml erlemeyer, bottom diameter of 6 cm and top diameter of 2.5 cm.
Outside circum. of 2*pie*r integrated from 1.25 to 3? I know I am missing something here.
http://mathworld.wolfram.com/ConicalFrustum.html
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15 Jun 10
1 edit
Originally posted by mlpriorfor your example I get
I think I found a solution:
http://mathworld.wolfram.com/ConicalFrustum.html
119*Pi/16
start by finding 2 points.
I used; A(0,3) & B(7/4,5/4)
Find the slope, which I get to be -1
Next find equation of line
y = 3-x
then since lateral surface area of a right circular cylinder = 2*Pi*r*h
r being f(x) and h = dx
we have
2*pi*Int{0, 7/4}[3-x]dx
= 2*pi*[ 3x-1/2x^2 |{0,7/4}]
= 119*pi/16 [mm^2]
EDIT: but according to wolfram im missing a factor of sqrt(2)!?( bummer)
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15 Jun 10
Originally posted by joe shmoI found where I went wrong. Apparently the formula for a surface of revolution is based off the frustrum of a cone ( the very thing you seek to find ) and not the right circular cylinder.
for your example I get
119*Pi/16
start by finding 2 points.
I used; A(0,3) & B(7/4,5/4)
Find the slope, which I get to be -1
Next find equation of line
y = 3-x
then since lateral surface area of a right circular cylinder = 2*Pi*r*h
r being f(x) and h = dx
we have
2*pi*Int{0, 7/4}[3-x]dx
= 2*pi*[ 3x-1/2x^2 |{0,7/4}]
= 119*pi/16 [mm^2]
EDIT: but according to wolfram im missing a factor of sqrt(2)!?( bummer)
S = 2*pi*[Int{a,b}r(x)*Sqrt( 1 + (r'(x))^2 )]
can anybody explain why it would need to be done this way?
Eric