Solve Integration: 4π∫x√(x2+2x+1)/4x

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Homework Help Overview

The discussion revolves around the integration of a function related to the surface area generated by revolving a loop defined by the equation 9y^2=x(3-x)^2 about the y-axis. The original poster seeks assistance with the integral involving the expression 4π∫x√(x²+2x+1)/4x.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the simplification of the integrand, noting that the expression x²+2x+1 can be rewritten as (x+1)². There are inquiries about the setup of the integral and whether it has been done correctly. Some participants express uncertainty about the constants involved in the integral.

Discussion Status

The discussion is ongoing, with participants providing feedback on the integral setup and expressing willingness to check the work. There is acknowledgment of the challenges associated with integrating functions that involve square roots, and some participants are exploring the implications of considering only the top half of the loop.

Contextual Notes

Participants mention that the integral is evaluated from 0 to 3, and there is a focus on ensuring the correct interpretation of the surface area problem, particularly regarding the presence of negative values in the context of the loop.

elitespart
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The loop 9y^2=x(3-x)^2 is revolved about the y-axis. Find the area of the surface generated this way.

I just need help with integration:

[tex]4\pi\int x\sqrt{\frac{x^{2}+2x+1}{4x}}[/tex]

I know I can pull out the 4 but after that I'm lost. Any suggestions?
 
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x^2+2*x+1=(x+1)^2. There's another thing you can pull out.
 
oh sweet. Thanks. oh if anyone has time to check could you please let me know if I set up the integral properly?
 
You know there is something special about problems like surface area and arc length? With that square root in them, you can almost never do them exactly. You have to set up a problem carefully to get it to have an exact solution. Since yours cracked so easily, I'm going to guess it's correct, just based on probability. But, sure, I'll check it.
 
Dick said:
You know there is something special about problems like surface area and arc length? With that square root in them, you can almost never do them exactly. You have to set up a problem carefully to get it to have an exact solution. Since yours cracked so easily, I'm going to guess it's correct, just based on probability. But, sure, I'll check it.

thanks. and I forgot to add that it's from 0 to 3. and you're right - I ended up with 60.9422...
 
Last edited:
It looks pretty good as far as the integral goes. Why do you have 4pi instead of 2pi. I might just be getting tired.
 
Dick said:
It looks pretty good as far as the integral goes. Why do you have 4pi instead of 2pi. I might just be getting tired.

Because I'm only considering the top half of the loop in the integral.
 
Oh. So there's negative y's as well. I was getting tired. Carry on.
 
lol no worries. Thank you again.
 

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