Question

What is the surface area when the arc of y = (e^(-x)), x >=0 is rotated about the x-axis?

What I intend to use is
S = 2pi integral(from 0 to (some number)) of (1/e^x)(1+ (-e^(-x))^2 dx

is this right? and what would be the 'some number' would that be infinity?

Answer 1

To hard wish I could help you

Answer 2

...Thanks though.

Answer 3

Revolution of a solid, but you're looking for surface area?

Answer 4

...yeah. theres actually 3 terms here... surface area, arc length and rotation. surface area and rotation are kinda tied.

Answer 5

Do you know how to use the ring method?

Answer 6

ring method...kinda. its prolly the cylindrical method i know.

Answer 7

ok ill try that.

Answer 8

I'll work on it. In the meantime, this might help with the concept.

http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

Answer 9

Your limits should be 0 to infinity, as the function converges. then if you use the general formula in the lecture, it should just be straight up integration from there.

Answer 10

okay thanks. apparently, it wasnt a good idea to bring up calculus here.

"If we wanted to we could also derive a similar formula for rotating on about the y-axis. This would give the following formula." <--- this is the formula i was blabbing about up in my question.

so I would write the equation then as:

S = 2pi int_0^(infinity) (1/e^x)(sqrt(1 + (-1/e^x)^2) dx

thanks

Answer 11

wait...

S = 2pi lim(t--> infinity) int_0^(t) (1/e^x)(sqrt(1 + (-1/e^x)^2) dx

Answer 12

I'm not sure about those limits. Hang on, I need to work this out

Answer 13

ok

Answer 14

Yeah, I think the equation we had before is right. Plug it into wolframalpha.com and you should be set

Answer 15

Night!

Answer 16

Ok thanks. Night