Asked by Alex
Find the area of the surface generated by revolving the curve y =
(x + 3)^1/2, 2 ≤ x ≤ 4
about the x-axis.
(x + 3)^1/2, 2 ≤ x ≤ 4
about the x-axis.
Answers
Answered by
TutorCat
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
Answered by
Alex
I tried using the formula there, but my answer seems wrong. I end up with the integral of radical x + a weird fraction, which doesn't seem to integrate well. Am I doing something wrong or is it just a complicated problem?
Thanks for the help
Thanks for the help
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