# [SOLVED] [Calculus]--Help with surface area of revolution please

• Mar 11th 2008, 05:41 PM
coe236
[SOLVED] [Calculus]--Help with surface area of revolution please
Problem: The curve y=e^(-x), x>0 is revolvd about the xaxis. Does the resulting surface have finite or infinite area? [Remember tht you can sometimes decide whether improper integral converges w/out calculating it exactly]

Solution Attempt:
Surface area over [a,b]= 2(pi)*integral[f(x)*sqrt(1+f'(x)^2)dx]
Comparison test for improper integrals assuming f(x)>g(x)>0 for x>a: if integral[f(x)dx] on [a,infinity] converges, then integral[g(x)dx] on [a,infinity] also converges.

So,
Surface area= 2(pi)*integral[e^(-x)*sqrt(1+e^(-2x))dx].. are the bounds [0,infinity]? What do I do after that to find out if it's finite or not? Do I use the comparison theorem? I'll appreciate any help. Thanks in advance
• Mar 11th 2008, 06:03 PM
mr fantastic
Quote:

Originally Posted by coe236
Problem: The curve y=e^(-x), x>0 is revolvd about the xaxis. Does the resulting surface have finite or infinite area? [Remember tht you can sometimes decide whether improper integral converges w/out calculating it exactly]

Solution Attempt:
Surface area over [a,b]= 2(pi)*integral[f(x)*sqrt(1+f'(x)^2)dx]
Comparison test for improper integrals assuming f(x)>g(x)>0 for x>a: if integral[f(x)dx] on [a,infinity] converges, then integral[g(x)dx] on [a,infinity] also converges.

So,
Surface area= 2(pi)*integral[e^(-x)*sqrt(1+e^(-2x))dx].. are the bounds [0,infinity]? What do I do after that to find out if it's finite or not? Do I use the comparison theorem? I'll appreciate any help. Thanks in advance

The integral limits are x = 0 to x = +oo.

To do the integral, make the substitution $\displaystyle u = e^{-x}$:

1. $\displaystyle dx = -e^{-x} \, du \Rightarrow dx = - \frac{du}{u}$.

2. $\displaystyle x = 0 \Rightarrow u = 1$ and $\displaystyle x = +\infty \Rightarrow u = 0$.

Therefore:

$\displaystyle S = 2 \pi \int_{1}^{0} u \, \sqrt{1 + u^2} \, \left( - \frac{du}{u} \right) = -2 \pi \int_{1}^{0} \sqrt{1 + u^2} \, du = 2 \pi \int_{0}^{1} \sqrt{1 + u^2} \, du$ ......
• Mar 11th 2008, 06:16 PM
coe236
Quote:

Originally Posted by mr fantastic
The integral limits are x = 0 to x = +oo.

To do the integral, make the substitution $\displaystyle u = e^{-x}$:

1. $\displaystyle dx = -e^{-x} \, du \Rightarrow dx = - \frac{du}{u}$.

2. $\displaystyle x = 0 \Rightarrow u = 1$ and $\displaystyle x = +\infty \Rightarrow u = 0$.

Therefore:

$\displaystyle S = - 2 \pi \int_{1}^{0} u \, \sqrt{1 + u^2} \, \left( - \frac{du}{u} \right) = -2 \pi \int_{1}^{0} \sqrt{1 + u^2} \, du = 2 \pi \int_{0}^{1} \sqrt{1 + u^2} \, du$ ......

Thanks very much--really quick--this might be a dumb question, but why is it negative 2(pi) instead of positive?
• Mar 11th 2008, 06:22 PM
mr fantastic
Quote:

Originally Posted by coe236
Thanks very much--really quick--this might be a dumb question, but why is it negative 2(pi) instead of positive?

Whoops. My mistake with cut and paste. I've edited my reply, correcting that error.
• Mar 11th 2008, 06:27 PM
coe236
Gotcha. Thanks again (Smile)