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Solution 1: The answer is$\frac{2b^5\pi}{5} \times \left( 1 -\frac{a}{\sqrt{1+a^2}}\right)$ Solution 2: I want to decide which solution is correct? Would you help me in this task?

So it would appear that in the first case you are calculating the integral of $r^2$ over $R$ and in the second the integral of $1$ over $R$ (the volume of $R$)
this is not a volume, just a triple integral $$\displaystyle a>0$$ and $$\displaystyle b>0$$ in spherical coordinates $$\displaystyle \int _0^{2\pi }\int _0^{\arctan (1/a)}\int _0^b\rho ^4\sin \phi d\rho d\phi d\theta =\frac{2}{5} \left(1-\frac{a}{\sqrt{1+a^2}}\right) b^5 \pi$$ in cylindrical coordinates $$\displaystyle \int _0^{2\pi }\int _0^{\frac{b}{\sqrt{a^2+1}}}\int _{a r}^{\sqrt{b^2-r^2}}\left(r^2+z^2\right)rdzdrd\theta =\frac{2}{5} \left(1-\frac{a}{\sqrt{1+a^2}}\right) b^5 \pi$$
WMDhamnekar said: View attachment 44608 Solution 1: View attachment 44610 The answer is$\frac{2b^5\pi}{5} \times \left( 1 -\frac{a}{\sqrt{1+a^2}}\right)$ Solution 2: View attachment 44611 I want to decide which solution is correct? Would you help me in this task? Click to expand... You can check this yourself. If I'm not mistaken (and I often am!), as $a\to 0$ the volume should go to that of a hemisphere of radius $b$ That is the term in front of the bracketed term should be $(2/3) \pi b^3$