Asked by Akitsuke
Find the area of the surface generated by revolving the given curve about the y-axis.
8xy^2=2y^6+1 , 1<=y<=2.
thank you so much.. :)
8xy^2=2y^6+1 , 1<=y<=2.
thank you so much.. :)
Answers
Answered by
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.
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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