# Thread: Surface area by revolving y=[2x^(3/2)]/3 - [x^(1/2)]/2 about the given axis.

1. ## Surface area by revolving y=[2x^(3/2)]/3 - [x^(1/2)]/2 about the given axis.

Hi. I am trying to solve the following problem without success.
Find the area of the surface generated by revolving the function y=[2x^(3/2)]/3 - [x^(1/2)]/2 between (0,0) and (9, 33/2), when the curve revolves around the y-axis.
I am stuck into not being able to change the equation from f(x) to f(y). My work ends at 36y^2=x(4x-3)^2.
(Calculus, Howard Anton, 3rd Edition, pg 417, exercise 25).
Thank you so much.

2. Originally Posted by denifelix
Hi. I am trying to solve the following problem without success.
Find the area of the surface generated by revolving the function y=[2x^(3/2)]/3 - [x^(1/2)]/2 between (0,0) and (9, 33/2), when the curve revolves around the y-axis.
I am stuck into not being able to change the equation from f(x) to f(y). My work ends at 36y^2=x(4x-3)^2.
(Calculus, Howard Anton, 3rd Edition, pg 417, exercise 25).
Try using what you learned back in algebra:

Did you do the graph of the original function, where y was in terms of x? If so, did you note the restrictions on the domain and range? If so, then what should you do after dividing through by 36?

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