I am asked to find the moment about the $x$ axis for a thin wire of constant density. This thin wire lies along the curve $y=\sqrt{x}$ and the limits for integration are $x=0$ and $x=2$.

I know from my textbook that the moment about the $x$ axis is: $M_y = \int \tilde{y} dm$

Because this is a thin wire, I know that I need to subdivide the wire into small segments for integration. I have the following for relevant data for each segment:

Length: $dl = \sqrt{x}dx$

mass: $dm = \delta dl = \delta \sqrt{x}dx$

It's the part about the distance of the center of mass to the $x$ axis that I think I'm missing. I have the following:

$\tilde{y} = \sqrt{x}$

Therefore, my final integral is:

$$ M_y = \int \tilde{y} dm = \int_0^2 \delta \sqrt{x} \sqrt{x} dx = \delta \int_0^2 x dx = \delta \left.\frac{1}{2}x^2\right|_0^2 = \delta 2 $$

This particular problem is an odd numbered problem and so I know that I've got it incorrect. Please help me to see where I'm going wrong.

Thanks, Andy


4 Answers 4


The center of mass for a wire of constant density making a curve $y=f(x)$ between $x=a$ and $x=b$ is

$$\bar{y} = \frac{\int_a^b dx \: y \sqrt{1+y'^2}}{\int_a^b dx \:\sqrt{1+y'^2}}$$

In the case you describe

$$\bar{y} = \frac{\displaystyle \int_0^2 dx \: \sqrt{x} \sqrt{1+\frac{1}{4 x}}}{\displaystyle \int_0^2 dx \: \sqrt{1+\frac{1}{4 x}}}$$

The top integral is relatively simple and is equal to

$$\int_0^2 dx \: \sqrt{x+\frac{1}{4}} = \frac{2}{3} \left ( \frac{27}{8} - \frac{1}{8} \right ) = \frac{13}{6}$$

The bottom integral may be evaluated by making the substitution $\sec{\theta} = \sqrt{1+\frac{1}{4 x}}$ to get

$$\int_0^2 dx \: \sqrt{1+\frac{1}{4 x}} = \frac{1}{2} \int_{\arccos{\sqrt{8}/3}}^{\pi/2} d\theta \: \csc^3{\theta} = \frac{1}{8} \left(12 \sqrt{2}+\log \left(17+12 \sqrt{2}\right)\right)$$

Therefore your center of mass is

$$\bar{y} = \frac{52/3}{12 \sqrt{2}+\log \left(17+12 \sqrt{2}\right)} \approx 0.846$$


The element of arclength is $ds=\sqrt {1+(\frac {dy}{dx})^2}dx$ or $ds=\sqrt {1+(\frac {dx}{dy})^2}dy$ and $dm=\delta ds$ To find the center of mass in the $y$ direction you need to find the average $y$. Please check your text, the center of mass in $y$ should be $y_{CM}=\frac 1m\int y dm$, then the moment of inertia in the $y$ direction around the center of mass is $\int (y-y_{CM})^2dm$. The expression you give doesn't have the proper units for MOI, which should be mass*length^2

  • $\begingroup$ The question/comment I have is really relevant to both responders so I'll put it here. This is apparently, one of the areas of misunderstanding that I had. The problem asked me to find the moment about the x-axis which I understood to be $\tilde{y}$. So this is what I was trying to find. Because the problem was worded thusly, I didn't think it was a question of "Find the Center of Mass." I thought that finding center of mass was to find $(\bar{x},\bar{y})$. At any rate, thank you both for you guidance. $\endgroup$ Feb 22, 2013 at 0:48

The top answer is wrong, it goes too far. The question isn't "find the center of mass of this area under the curve" its "find the moment of this wire about the x-axis" and the answer is 13/6 (times density) as he calculates in the top integral. (This question is from a Thomas Calculus book.) I'm not allowed to comment yet on Ron's post because I just joined to fix this.

The actual equation (from Ron's post):

$$\int_a^b dx \: y \sqrt{1+y'^2}$$

For this particular problem:

$$\int_0^2 dx \: \sqrt{x+\frac{1}{4}} = \frac{2}{3} \left ( \frac{27}{8} - \frac{1}{8} \right ) = \frac{13}{6}$$

So the actual answer is just the top integral Ron Gordon set up, which is basically the integral of (function * arc length * density).

  • $\begingroup$ So now what is $\bar y$? $\endgroup$
    – Narasimham
    Jul 30, 2020 at 12:17
  • $\begingroup$ @Narasimham I do not know. As the original poster said, this is a question from a calculus book, the answer is in the back, the answer is 13/6. I went on an internet search to find if someone had asked how it was calculated because I couldn't figure it out and realized Ron's answer was right - at least the top half was. $\endgroup$
    – user242506
    Aug 8, 2020 at 1:52

For arc-length integration

$$\bar{y} = \frac{\int \: y \sqrt{1+y'^2}\,dx }{\int \sqrt{1+y'^2}\;dx}=\frac{Nr}{Dr}$$

We can evaluate Numerator and Denominator by parametrization easier with $t$, max value being $t_m= y_m/(2f)$.

$$ y^2= 4 f x; (f=\frac14\;); x=ft^2,y=2ft, x'= 2ft\;y'=2f;$$

Nr=$\int 2ft\;\sqrt{1+4f^2}\;2ft\;dt= 4f^2\sqrt{1+4f^2}\;t_m^3/3$

Dr=$\int \sqrt{1+4f^2}\;2ft\;dt=f\sqrt{1+4f^2}\;t_m^2\;$

Divide,simplify $\bar y= \dfrac{4f}{3}t_m=\dfrac{2}{3}y_m= \dfrac{2\sqrt2}{3}\approx 0.942809 $


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