Integral Calculus
posted by Akitsuke .
Find the area of the surface generated by revolving the given curve about the yaxis.
8xy^2=2y^6+1 , 1<=y<=2.
thank you so much.. :)

Integral Calculus 
drwls
You need x as a function of y to do the integration.
In this case, x(y) = 1/(8y^2) + (y^4)/4
To get the surface area, evaluate:
Integral of 2 pi x(y) dy
(y = 1 to 2)
The indefinite integral is
2 pi [1/(24y^3) + y^5/20]
= pi[(y^5)/10  1/(24y^3)]
Evaluate it at y=2 and y=1, and take the difference. Check my math.
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