# Finding the Surface Area of the curve?

• July 25th 2010, 04:19 PM
racewithferrari
Finding the Surface Area of the curve?
Finding the Surface Area of the curve?

y = 1/x
1 ≤ x ≤ ∞ about x-axis.

Thanks
• July 25th 2010, 05:19 PM
I didn't check through your integration, but when you have an improper integral $\int_a^{\infty}f(x)dx$, just compute the limit $\lim_{b->\infty}\int_a^bf(x)dx$

Are you familiar with improper integrals?
• July 25th 2010, 05:42 PM
racewithferrari
After solving it, the answer come to ∞. Right?
• July 25th 2010, 05:59 PM
Quote:

Originally Posted by racewithferrari
After solving it, the answer come to ∞. Right?

Actually, I think you have an error in your integration. I checked it, and there's a problem:

$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}dx$

Add the fractions in the radical $\sqrt{\frac{x^4+1}{x^4}}=\frac{\sqrt{1+x^4}}{x^2}$

Plug this into the integral to get $\int \frac{\sqrt{1+x^4}}{x^3}dx$.
• July 25th 2010, 06:29 PM
skeeter
$\displaystyle \int_1^{\infty} \frac{\sqrt{1+x^4}}{x^3} \, dx > \int_1^{\infty} \frac{1}{x} \, dx$

note that it is rather easy to show that $\displaystyle \int_1^{\infty} \frac{1}{x} \, dx$ diverges.
• July 26th 2010, 01:34 AM
racewithferrari
I did nothing wrong. I combined num and dem. The sqrt sign is on the top only.