# Math Help - Z-coordinate of centroid , centre of mass prob

1. ## Z-coordinate of centroid , centre of mass prob

Hi!

Problem: Determine the z-coordinate of the centroid of the solid generated by revolving the quarter-circular area thorugh 90 degress about the z-axis.

Solution attempt:

I tried to calculate the volume of the solid. I got it to be $\frac{\pi a^{3}}{24}(4+3\pi)$

I have some trouble setting up the sum of moments integral, that is $\int z_{c} \; dz = V\bar{z}$

I tried to use a horizontal strip as differential element:
Witdh: $\sqrt{a^{2}-z^{2}}$

Height: $dz$

Moment arm: $z$

But I also need thickness, right? expressed in terms of z

Thx!

2. First , calculate the numinator : $I = {\int\int\int}_{D} z dV$

Since it is uniform , we may neglect the density .

$I = \int_0^{\frac{\pi}{2}} \int_a^{2a} \frac{ 4a^2 - r^2 }{2} rdr d\theta$

$= \frac{\pi}{4} [ 2a^2r^2 - \frac{r^4}{4} ]_{a}^{2a}$

$= \frac{\pi}{4} \frac{9a^4}{4}$

$= \frac{9a^4 \pi }{16}$

Oh i made a mistake , i thought that is a part of a sphere ...

The function should be $z = \sqrt{ a^2 - ( r-a)^2}$

so $I = \frac{\pi}{4}[ \int_a^{2a} (a^2 r - r(r-a)^2) dr ]$

$= \frac{\pi}{4} (a^4) [ \frac{11}{12} ]$

$= \frac{ 11 \pi a^4 }{48}$

and $V = \frac{\pi}{4} [\int_a^{2a} \sqrt{ a^2 - (r-a)^2} rdr$

Sub $r-a = a\sin{\theta}$ $\implies$

$V = \frac{\pi}{2} [ \int_0^{\frac{\pi}{2}} a^3 ( 1 + \sin{\theta}) \cos^2{\theta} d\theta ]$

$= \frac{\pi}{2} (a^3) [\frac{\pi}{4} + \frac{1}{3} ]$

$= \frac{a^3 \pi}{2} \frac{3\pi + 4}{12}$

$= \frac{ a^3 \pi ( 3\pi + 4) }{24}$

So bar-z $= \frac{11a}{2(3\pi + 4)}$

3. I dont get it, did you change to polar coordinates? Why do yo multiply by another r?

What is the moment arm?

4. Originally Posted by Twig
I dont get it, did you change to polar coordinates?
Well, yes. The " $r dr d\theta$ is kind of a tip off, don't you think? There is plenty of circular symmetry here so using polar coordinates is kind of obvious.

5. Originally Posted by HallsofIvy
Well, yes. The " $r dr d\theta$ is kind of a tip off, don't you think? There is plenty of circular symmetry here so using polar coordinates is kind of obvious.

Yes, true.

$z^{2}+(r-a)^{2}=a^{2}$ , it feels like he just changes x to r.

I am with on the moment arm: $\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}$

Shoudlnt it work just integrating "as usual" without polar coordinates, even if polar will make it easier?

What is wrong below?

$\int_{a}^{2a}\underbrace{\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}}_{z_{c}}\cdot \underbrace{\sqrt{a^{2}-(x-a)^{2}}}_{Height} \; dx$

I thought polar was $x=r\cdot cos(\theta)$

6. Originally Posted by Twig
Yes, true.

$z^{2}+(r-a)^{2}=a^{2}$ , it feels like he just changes x to r.

I am with on the moment arm: $\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}$

Shoudlnt it work just integrating "as usual" without polar coordinates, even if polar will make it easier?

What is wrong below?

$\int_{a}^{2a}\underbrace{\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}}_{z_{c}}\cdot \underbrace{\sqrt{a^{2}-(x-a)^{2}}}_{Height} \; dx$

I thought polar was $x=r\cdot cos(\theta)$
I think $\int_a^{2a} \sqrt{ a^2 - (x-a)^2 } ^2 ~dx$ is equal to the moment sum of circular plate with radius a and center (a,0)

$\int\int z dzdL ~~$ and $\int\int\int z dz dA$
are actually different .

The moment sum of the required solid is
${\int \int \int}_D z dz dA$
The reason using polar co. is for convenient calculation $dA = rdrd\theta$

So the moment sum is also equal to

${\int \int \int}_D z dz (rdrd\theta)$ $D = [ (\theta,r,z) \in {\mathbb{R}}^3 |0 \leq z \leq \sqrt{a^2 - (r-a)^2 } , a \leq r \leq 2a , 0 \leq \theta \leq \frac{\pi}{2}]$