- #1
think4432
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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 I am 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!
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 I am 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!