# I Surface Area of Volume of Revolution

1. Oct 10, 2017

### cphill29

The problem is, find the surface area of the volume of revolution generated by rotating the curve y=e2x between x=0 and x=2 about the x-axis.

Here's what I have so far...

SA=∫y√(1+y2)dx
=∫e2x√(1+4e4x)dx

and from here I'm not really sure what to do. Any help would be appreciated.

2. Oct 10, 2017

### Math_QED

You want to solve the integral $I = \int e^{2x} \sqrt{1+4e^{4x}}dx = \int e^{2x} \sqrt{1+(2e^{2x})²}dx$

To do this, I suggest you perform the substitution $u = 2e^{2x}$. The integral you then obtain can be easily solved using an appropriate trigonometric substitution.

Last edited: Oct 10, 2017
3. Oct 10, 2017

### cphill29

OK, so setting u=2e2x, I get du=4e2x dx, and dx=1/4e2x du. Substituting that in, I get...

I=2π∫(e2x)(1/4)(1/e2x)√(1+u2) and then when I simplify...

I=1/2π∫√(1+u2) using a trig sub,

I=1/2π∫sec3θ dθ

Is this correct?

4. Oct 10, 2017

### Math_QED

The result you obtain is correct.

5. Oct 10, 2017

### cphill29

Thank you very much for your help!

6. Oct 10, 2017

### Math_QED

Little disclaimer, I just helped you solve the integral. I don't know the formula of revolution by heart, so I don't know whether you have that part of the question correct. Also, you should post questions like these in the homework sections ("calculus and beyond").