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$$\int_{0}^{2} \mathrm{~d} z \int_{0}^{\left(2 z-z^{2}\right)^{\frac{1}{2}}} \mathrm{~d} y \int_{0}^{\left(2 z-z^{2}-y^{2}\right)^{\frac{1}{2}}}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}} \mathrm{~d} x$$ By $$\int \frac{\mathrm{d} x}{\sqrt{x^{2}+a^{2}}}=\ln \left(x+\sqrt{x^{2}+a^{2}}\right)+C $$ $\Rightarrow$ $$\int_{0}^{2} \mathrm{~d} z \int_{0}^{\left(2 z-z^{2}\right)^{\frac{1}{2}}}\ln \left(x +\sqrt {x^2+y^2+z^2}\right) \bigg|_{0}^{\left(2 z-z^{2}-y^{2}\right)^{\frac{1}{2}}} \mathrm{~d} y $$ And I have no idea how to do next. If I consider the spherical coordinates, how can I determine value of $r \ \varphi \ \theta$

jhbihb
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    Continuing this integral is a little maddening (in the insane way). You should be recognizing spherical coordinates – Ninad Munshi Nov 11 '22 at 14:43
  • Mathematica: $\frac{\pi}{3}$. – David G. Stork Nov 11 '22 at 14:55
  • @NinadMunshi then how to determine the value by spherical coordinates – jhbihb Nov 12 '22 at 06:33
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    You have a typo/slip in your last equation: the first $x^2$ should be $x$. Remark: does it help to look at $x^2+y^2+(z-1)^2=1$? – ancient mathematician Nov 12 '22 at 07:51
  • @ancientmathematician thank you. It seems to make sense, but why is z integrated from 0 to 2, if so, z also seems intractable after coordinate transformation. – jhbihb Nov 12 '22 at 08:11
  • @ancientmathematician yes,it is 0 to 2 after I draw an image. Thank you! How can you recognize it? how can I determine value of $r \ \varphi \ \theta$ – jhbihb Nov 12 '22 at 08:28
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    I think we are just working out the potential of a uniform ball of radius $1$ at a point on the surface, so I'd expect to get $\frac43\pi$ which doesn't match @DavidG.Stork 's answer. But maybe we're only integrating over a quarter of the sphere - it's too long since I did this sort of stuff. – ancient mathematician Nov 12 '22 at 10:39

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From your bounds we know two things: the integral is entirely within the first octant, and we are dealing with some kind of sphere. The equation of surface can be found by dealing with the innermost $x$ bound:

$$x = (2z-z^2-y^2)^{\frac{1}{2}} \implies x^2+y^2+z^2=2z$$

which is a sphere of radius $1$ centered at $(0,0,1)$ (why?). To convert to spherical coordinates, simply plug into this expression

$$r^2 = 2r\cos\theta \implies \theta = \cos^{-1}\left(\frac{r}{2}\right)$$

which gives us the the bounds

$$\int_0^\frac{\pi}{2} \int_0^2 \int_0^{\cos^{-1}\left(\frac{r}{2}\right)} \frac{1}{r} \cdot r^2 \sin\theta \: d\theta \:dr\:d\varphi$$

$$=\frac{\pi}{2} \int_0^2 r-\frac{r^2}{2}\:dr = \frac{\pi}{3}$$

Ninad Munshi
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