Surface Area by rotating a curve

[xo.Stardust]

1. Find the surface area generated by rotating the curve about the y-axis

2. X=(1/3)Y^(3/2)-Y^(1/2) between 1 and 3

3. x' = (1/2)Y^(1/2) - (1/2)Y^(-1/2)
x'^2 = (y^2-2y+1)/4y
I'm not sure if that derivative is correct or if implicit differentiation is required. I also can't get any further using the surface area formula. Any help would be appreciated.

 

[Bohrok]

Your math is correct; you shouldn't need to use implicit differentiation. Which formula are you using for surface area?

 

[xo.Stardust]

https://www.physicsforums.com/latex_images/23/2361260-0.png

After substituting my data into the equation I can't simplify it far enough to take the integral

 

[Bohrok]

The first x in that integral should be a y.

Combine the 1 with (y^2-2y+1)/4y and try to get the quadratic part into (y + something)² so you can take its square root.

 

[xo.Stardust]

Combine the 1 with (y^2-2y+1)/4y and try to get the quadratic part into (y + something)² so you can take its square root.

I didn't think to do that, I think I've got it now. Thanks!