# Surface Area by rotating a curve

1. Sep 22, 2009

### 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.

2. Sep 22, 2009

### Bohrok

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

3. Sep 22, 2009

### xo.Stardust

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

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

Last edited by a moderator: May 4, 2017
4. Sep 22, 2009

### 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.

5. Sep 22, 2009

### xo.Stardust

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