- 14 Jun '10 21:36

I think I found a solution:*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.

http://mathworld.wolfram.com/ConicalFrustum.html - 15 Jun '10 03:54 / 1 edit

for your example I get*Originally posted by mlprior***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) - 15 Jun '10 15:43

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.*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)

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