surface area of an erlenmeyer flask

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.

Originally posted by mlprior
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.

I think I found a solution:

http://mathworld.wolfram.com/ConicalFrustum.html

Originally posted by mlprior
I think I found a solution:

http://mathworld.wolfram.com/ConicalFrustum.html

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)

Originally posted by joe shmo
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)

I 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.

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