Surface Area of Solid: f(x)=1/x, [1,∞) about x-axis

[MrShickadance]

Homework Statement

f(x) = 1/x
Interval [1, ∞) about the x-axis

Set-up the integral for the surface area of the solid

Then use the substitution u = x² and integrate using the formula:

∫ sqrt(u² + a²) / u² du = ln(u + sqrt(u² + a²) - sqrt(u² + a²) / u + C
a is a constant

Homework Equations

S = 2pi * ∫ (f(x) * sqrt(1 + [f`(x)]²) dx from a to b

The Attempt at a Solution

First, I found the derivative of (1/x) which is -1/x²

I then plugged f(x) and f`(x) into the surface area equation

I squared f`(x) to get (1/x⁴)

My equation is 2pi ∫ (1/x) * sqrt(1 + (1/x⁴) from 1 to infinity of course, which I will change to the limit as b approaches infinity because it is an improper integral.

I simplified the fractions under the radical to get sqrt((x⁴ + 1) / x⁴)

I took the square root of the denominator to get x²

Lastly, I multiplied (1/x) by sqrt(x⁴ + 1) / x² to get
sqrt(x⁴ + 1) / x³

If u = x² then this is not in the correct form to use the formula that was given to me.

How can I get the denominator to equal x⁴?

I will figure out the rest of the problem from there.

Here is my written attempt:

Thanks!

 

[vela]

That's what you get for being sloppy and leaving the dx out of the original integral.

 

[HallsofIvy]

Vela's point is that you have simply put the "x" integrand into terms of "u" and appended "du". In fact, if you had included "dx" in the original integral you would have realized that, since $u= x^2$, $du= 2x dx$.

 

[MrShickadance]

Thanks Halls. I think I have arrived at the correct answer now. Unfortunately, Wolfram doesn't seem to want to load the answer. It's stuck loading and I've tried refreshing the page. Was it really that difficult? haha