# Thread: Calc 2- Surface Area of Parametric Curves

1. ## Calc 2- Surface Area of Parametric Curves

Hey guys, I'm just stuck on the manipulation part of this problem but I'll set it up for ya'll.

Y=√(2x-x^2)
1/2≤x≤3/2
Rotating around the X-Axis

I set up the problem and plugged it into the equation which I believe is,

S=∫2πy√(1+(dy/dx))dx

When I plug in the derivative of Y, (1-x)/√(2x-x^2) under the square root in the main equation, I get lost as what to do next. I tried a few things but I'm either missing something or just flat out messing up my manipulation.

Thanks for your help! Audio

2. Bump!

3. Hi Audio,

I think your mistake is that you are using:
$S = \int 2 \pi y \sqrt{1+dy/dx} dx$

whereas you should be using:
$S = \int 2 \pi y \sqrt{1+(dy/dx)^2} dx$

I worked it through and it came out quite easily as $2 \pi$

Let me know if you are stuck!