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Let $S$ be the hemisphere with radius $1$ and centre along the $z$-axis in such a way that the south-pole of the hemisphere is located at the origin.

The mass function is $p(x,y,z) = 1-z$

I know that a sphere of radius 1 is given by the formula $x^2+y^2+z^2=1$. Knowing this I can solve for $z = \pm\sqrt{1-x^2-y^2}$, and then add +1 to lift the sphere up to $z=1$. $z= 1- \sqrt{1-x^2-y^2}$,

I can then insert for $z$ into the mass function. $$\iint 1-\left(1- \sqrt{1-x^2-y^2}\right) dA. $$

Further I can convert it into polar coordinates, $$\int_0^{2\pi}\int_0^{1}r\sqrt{1-r^2}dr d\theta.$$ I end up with the answer $\frac{2\pi}{3}$.

Is this correct?

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  • $\begingroup$ The circle is not centered at (0,0), look at your picture we do not have r=1. $\endgroup$ Dec 3, 2016 at 17:54

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It is not correct. Since at height $z$ we have a disc of radius $\sqrt{1-(1-z)^2}$ and constant density $(1-z)$, it follows, by integrating by sections, that the mass of the hemisphere is $$\int_{0}^1(1-z)\cdot \pi(1-(1-z)^2) dz=\frac{\pi}{4}.$$ The same result can be obtained following your approach: $$\int_{\theta=0}^{2\pi}\int_{r=0}^1\int_{z=1-\sqrt{1-r^2}}^{1}(1-z)\, (dz\, rdr\, d\theta).$$

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  • $\begingroup$ I was torn between doing what you did at the bottom, and what I did. I argued with myself that when you have a function and a double integral, you will get volume. Therefore I thought that if I had a mass function I would get the mass of a volume, but that isn't correct? If i have double integral, and a function for mass I will get the mass of an area (a thin plate)? $\endgroup$
    – David Lund
    Dec 3, 2016 at 18:15
  • $\begingroup$ @David Lund I am not sure that I have understood your question. Anyway, here you have to integrate the mass function over a solid so you have to use a triple integral. $\endgroup$
    – Robert Z
    Dec 3, 2016 at 18:19
  • $\begingroup$ @David Lund Any further doubt? $\endgroup$
    – Robert Z
    Dec 3, 2016 at 18:28
  • $\begingroup$ I meant, a double integral with a function inside of it gives volume, so I thought I could just insert it straight into the formula. And get the mass of the volume but I guess that isn't the case. If I have double integral (with mass a function) that just gives me the mass of a thin plate (area) in the double integral right? so that is why It wouldn't work. $\endgroup$
    – David Lund
    Dec 3, 2016 at 18:29
  • $\begingroup$ @David Lund Ah. Ok. I confirm that "a double integral with a function inside of it gives volume" can not be generalized in that way. $\endgroup$
    – Robert Z
    Dec 3, 2016 at 18:47

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