1. Find the area of the surface generated by revolving y = (sqrt)(2x-x^2), x = [0,1](adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Homework Help: Calculus II: Surface area revolving around an axis!

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