- #1

kscribble

- 10

- 0

Now I understand how the ACTUAL integral works to find the surface area, but I'm wondering why a different integral wouldn't work...

[tex]\int[/tex] 2[tex]\pi[/tex]*f(x) dx

wouldn't this add up the differential circumferences, so shouldn't this equation work? why do you have to use the Pythagorean theorem integral version to find the surface area?

thanks if you can answer this question :)