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

[Oblakastouf]

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]

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

 

[Oblakastouf]

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

 

[Dick]

It's kind of a long haul. You start by integrating by parts u=sec(t), dv=sec(t)^2*dt. It probably goes a little easier if you go back to the integral of sqrt(1+x^2)*dx and substitute x=sinh(u), if you are ok with hyperbolic functions.

 

[Oblakastouf]

It's kind of a long haul. You start by integrating by parts u=sec(t), dv=sec(t)^2*dt. It probably goes a little easier if you go back to the integral of sqrt(1+x^2)*dx and substitute x=sinh(u), if you are ok with hyperbolic functions.

I am, but I'm in a class that doesn't use them yet lol.