# Centre of mass of a solid hemisphere. What am I doing wrong?

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1. May 28, 2015

### Marvin94

I have to calculate the centre of mass of the drawn hemisphere. I don't understand where I make mistakes. The process I followed on the above image is the following:

(I) Here I simply translated position vector components from cartesian to polar coordinates.
(II) Formula of x-component of position vector of centre of gravity (this position vector clearly lies on the x-axis, so I'm interested only in this component)
(III) This is the volume element for polar coordinates system.

After that, I just put (III) into (II).

The problem is that, the result of the last (triple) integral (computable easily with wolfram alpha) seems to result zero! I thought that the mistake could be in the integration limits, but they look to be correct. Where am I doing wrong?

Thank you very much in advance.

2. May 28, 2015

### BvU

Hi Marv,

Check your integration limits anyway. Looks to me that you aren't integrating over the righthand half of the sphere....

3. May 28, 2015

### Marvin94

I tried to understand why what I wrote should be wrong, but I don't see the mistake. Which limits do you think are correct?

4. May 28, 2015

### BvU

Well, you havent shown what $\theta$ ande $\phi$ are, so I can't tell, but you can !
 but it's implicit in your coordinate transformation. Check if Rx > 0 always during the integration...

5. May 28, 2015

### Marvin94

θ = angle between position vector and x-axis
ϕ = angle between position vector and z-axis

6. May 28, 2015

### Marvin94

I used an ambiguous notation: I should precise, that the initial vector r (see (I) ) is a general position vector. The vector rx I use after that refers to the position vector of the centre of mass (not the general one!).

7. May 28, 2015

### BvU

Check if your rx is always positive.
Check if your ry really runs from -1 to +1
Check if your rz really runs from -1 to +1

As an exercise, do the integration yourself, instead of with the Wolfram crutches.

8. May 28, 2015

### BvU

Nope. Not if $z = r\cos \theta$

9. May 28, 2015

### theodoros.mihos

Check terms in $$r_{cm} = \frac{1}{V}\int\int\int \,r\,dV$$

10. May 28, 2015

### Marvin94

Thanks you all a lot :)