# Surface Area of y = e^5x revolved around the x-axis from 0 to ln(4)

## Homework Statement

http://i47.tinypic.com/1z6naa.jpg

Note... I used wolfram alpha to get the answer, I did not get it myself... So I still need help. The answer shown is correct, so you'll know if you got it.

## Homework Equations

Integral [0, ln(4)] sqrt(1+(dy/dx)^2)

## The Attempt at a Solution

2pi Integral [0, ln(4)] y*sqrt(1+(dy/dx)^2)

2pi Integral [0, ln(4)] (e^5x)*sqrt(1+5e^5x^2)dx

u = 5e^5x
du = 25e^5x dx
dx = du/25e^5x

2pi Integral [0, ln(4)] (e^5x)*sqrt(1+u^2)du/25e^5x

2pi Integral [0, ln(4)] sqrt(1+u^2)du

u = tan(t)

2pi/25 Integral [0, ln(4)] sqrt(1+tan^2(t))du

2pi/25 Integral [0, ln(4)] sqrt(sec^2(t))du

2pi/25 Integral [0, ln(4)] sec(t)du

du = sec^2(t)dt
dt = du*cos^2(t)

2pi/25 Integral [0, ln(4)] cos^2(t)/cos(t)dt

2pi/25 Integral [0, ln(4)] cos(t)dt

Edit bounds...

[arctan(5), arctan(5e^(5*ln(4)))]

Then get ****ed over with an answer of .0048...

What did I do wrong.

Dick
Homework Helper
If du=sec^2(t)*dt, then sec(t)*du is sec^3(t)*dt.

Right... My mistake, but I'm also having trouble with integration, and that isn't my strong suit, how would I integrate that?

Dick