I Surface Area of Volume of Revolution

[cphill29]

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

Here's what I have so far...

SA=∫y√(1+y²)dx
=∫e^2x√(1+4e^4x)dx

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

 

[member 587159]

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.

 

[cphill29]

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

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

I=1/2π∫√(1+u²) using a trig sub,

I=1/2π∫sec³θ dθ

Is this correct?

 

[member 587159]

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

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

I=1/2π∫√(1+u²) using a trig sub,

I=1/2π∫sec³θ dθ

Is this correct?

The result you obtain is correct.

 

[cphill29]

Thank you very much for your help!

 

[member 587159]

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").