Calculus II: Surface area revolving around an axis!

  • Thread starter think4432
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  • #1
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1. Find the area of the surface generated by revolving y = (sqrt)(2x-x^2), x = [0,1]

a. About the x axis

b. About the y-axis

This problem is from the section about surface area revolving around an axis.

I actually did the problem but im not sure on my answer.

For around the x axis I got an answer of 2pi. The integral that I set up was integrating 2pi as a constant and then dx from 0 to 1.

For the b part I got the integral of 2pi as a constant and then integrating x / (sqrt)(2x-x^2)

Please see if this is correct, if not please explain to me how it should be done.

Thank you!
 

Answers and Replies

  • #2
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I agree with you on both accounts.
 
  • #3
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So this is correct?

Wow. I thought I had done something completely wrong because I got a answer such as 2pi.

But thank you! :]
 
  • #4
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I agree with you on both accounts.

Are you sure on this?

Because on the online homework its showing me that it is incorrect?

Maybe its a mistake by the professor?

Please check again? Or can someone else please help?
 
  • #5
370
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Are you sure on this?

Because on the online homework its showing me that it is incorrect?

Maybe its a mistake by the professor?

Please check again? Or can someone else please help?

Yep I'm pretty sure, i've just double checked. If you've written down the right question the answers to both of your questions were correct. What was given as the answer out of interest?
 
  • #6
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Yep I'm pretty sure, i've just double checked. If you've written down the right question the answers to both of your questions were correct. What was given as the answer out of interest?

Oh! The program online is just really weird...and wanted me to put the answer in 2* the actual value of pi.

It didn't show me the correct answer right away...so I had to keep trying.

It wanted the answer in a decimal form.

Thank you very much for all your help!

Greatly appreciated!
 

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