The answer is radius/2, but I am confused as to why one of my approaches didn't work.


Let radius = a.
Let r^2 = a^2 - z^2 from z = 0 to z = a
Therefore, the surface area of such a hemisphere would be 2*pi*a^2


I will compare two methods, both single integration. They both are exploiting the fact that the COM of a circle is at the middle.






--- the integral of 2pi*z*sqrt(a^2-z^2) with respect to z from 0 to a - Wolfram|Alpha/2*pi*a^2


= a/3




--- the integral of 2pi*z*sqrt(a^2-z^2)*((a^2)/(a^2-z^2)) with respect to z from 0 to a - Wolfram|Alpha/2*pi*a^2



= a/2


[Note: What this one does is the calculus of the change in arc-length along the circle rather than the calculus of the change in z. What it does is it replaces dz with sqrt((dz)^2 + (dr)^2 )]





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Now what I don't understand is why these two get different answers. Whenever I've integrated with dz before in a sphere like this I usually get the right answer. Is this my incomplete understanding of limits? My teacher said that something weird happens with the first approach as z approaches a (the two infinities clash or something).

I've also tried finding the center of mass of a solid hemisphere both ways and the standard dz one works (3a/8) and the arc-length one doesn't (3a/4)

Also sorry about not using LATEX. I couldn't get it to work.


Thanks for any responses!


**
tl;dr why integrate by arc-length?**