Finding the Surface Area of the curve?

y = 1/x

1 ≤ x ≤ ∞about x-axis.

Thanks

Printable View

- Jul 25th 2010, 04:19 PMracewithferrariFinding the Surface Area of the curve?
Finding the Surface Area of the curve?

**y = 1/x**

**1 ≤ x ≤ ∞**about x-axis.

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

Are you familiar with improper integrals? - Jul 25th 2010, 05:42 PMracewithferrari
After solving it, the answer come to ∞. Right?

- Jul 25th 2010, 05:59 PMadkinsjr

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

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

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

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

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