# Thread: First moment of arc

1. ## First moment of arc

Find the first moment, about the y-axis, of the arc of the parabola $\displaystyle y^2=1-x$ for which $\displaystyle x\geq o$. Find also the length if the arc and hence find the centroid of the arc. Use Pappus' first theorem to find the area of the surface generated when the arc rotates completely around the y-axis.

Moments is one of the things i don't exactly understand that well. I am given the formula $\displaystyle M=\sum y \delta x$ for the moment about the x-axis.
So I concluded that $\displaystyle M=\sum x\delta y$ for the moment about the y-axis.
$\displaystyle M=\displaystyle \int ^1_{-1} (1-y^2) \sqrt{1+4y^2} dy$
Solving this, I get: $\displaystyle \frac{7\sqrt{5}}{16}+\frac{17}{32}arsinh 2$. Answers given for this question are: 2; $\displaystyle \pi$' $\displaystyle \frac{2}{\pi}$; $\displaystyle 4\pi$.
Thanks!

2. What you have got is the moment for the two dimensional region under the arc, not the one-dimensional arc itself. The differential of areclength with x= f(y) is $\displaystyle \sqrt{1+ f'^2(y)}dy$. You need to integrate $\displaystyle \int_{-1}^1 x\sqrt{1- f'^2(y)}dy$.

3. is it $\displaystyle \int_{-1}^1 x\sqrt{1- f'^2(y)}dy$ or $\displaystyle \int_{-1}^1 x\sqrt{1+ f'^2(y)}dy$?
I did do $\displaystyle \int_{-1}^1 x\sqrt{1+ f'^2(y)}dy$. and the answer i showed was what i got, sorry for the error, i missed typing the last part of the equation.
Thanks