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

**coe236** · March 11th 2008, 05:41 PM

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

**mr fantastic** · March 11th 2008, 06:03 PM

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 $u = e^{-x}$ :

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

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

Therefore:

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

**coe236** · March 11th 2008, 06:16 PM

Originally Posted by mr fantastic

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

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

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

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

Therefore:

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

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

**mr fantastic** · March 11th 2008, 06:22 PM

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.

**coe236** · March 11th 2008, 06:27 PM

Gotcha. Thanks again

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