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

• 12-06-2017, 10:20 PM
denifelix
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.
• 12-07-2017, 06:25 PM
stapel
Quote:

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? ;)
• 12-12-2017, 10:21 PM
denifelix
Thank you
Quote:

Originally Posted by stapel
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? ;)

Hi and thank you for taking the time to check on this. I did graph the original function, and did notice that it is negative for y between 0 and 0.75 (perhaps that's what you refered to). Here is the bad news. I could not go beyond what you told me, namely dividing by 36, and maybe square root everything. I know it sounds like I totally don't know what I am doing. Here is the good news. I went back to the chapter and saw how the formula for the surface was figured out, and applied the same principle for y-axis rotation, instead of x-axis, without changing the function from f(x) to f(y). The formula ended up as:

S=(Integral between a and b)2pi(x){1+[f'(x)]^2}^1/2 dx

So, instead of using f(x) as the radius component, I used x, and got the answer in the back of the book. I am still not sure if this is mathematically correct, or if that is how the exercise was intended to be resolved. In any case, thank you for your interested. Would you have any comments about this substitution from f(x) to x, regarding x-axis to y-axis rotation without changing the function from f(x) to f(y)? Thank you again. DF