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title : How to integrate this function?(substitution)

$$\iiint_{x^2+y^2+z^2\leq 4} \frac{x}{\sqrt{(x-1)^2 + (y-2)^2+(z-2)^2}}dxdxydz$$


(hint : integration by substitution )

I'm tried this using method by integration by substitution by sphere coordinate system But failed. Please help.

p.s.) If someone put the substitute successfully, Please tell me the reason or principle why should I put those like that. :)

Thanks.

The answer was $\frac{128\pi}{405}$

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Hmm... I don't get your answer. Edit: made a mistake in the intgration limit.

We first rotate so that $(x,y,z)=(1,2,2)$ is on the positive $z'$-axis: $$ \begin{pmatrix} x'\\y'\\z' \end{pmatrix} = \begin{pmatrix} \frac4{3\sqrt2}&-\frac1{3\sqrt2}&-\frac1{3\sqrt2}\\ 0&\frac1{\sqrt2}&-\frac1{\sqrt2}\\ \frac13&\frac23&\frac23 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} $$ So $$ I= \iiint_{x'^2+y'^2+z'^2\leq 4} \frac{\frac4{3\sqrt2}x'+\frac13z'}{\sqrt{x'^2+y'^2+(z'-3)^2}}\,\mathrm{d}x'\,\mathrm{d}y'\,\mathrm{d}z' $$ By symmetry, $$ \iiint_{x'^2+y'^2+z'^2\leq 4} \frac{x'}{\sqrt{x'^2+y'^2+(z'-3)^2}}\,\mathrm{d}x'\,\mathrm{d}y'\,\mathrm{d}z'=0 $$ so $$ I= \iiint_{x'^2+y'^2+z'^2\leq 4} \frac{\frac13z'}{\sqrt{x'^2+y'^2+(z'-3)^2}}\,\mathrm{d}x'\,\mathrm{d}y'\,\mathrm{d}z' $$ Now change to cylindrical polars: \begin{align*} I&=2\pi \int_{-2}^2\int_0^{\sqrt{4-z'^2}} \frac{\frac13z'}{\sqrt{\rho^2+(z'-3)^2}}\rho\,\mathrm{d}\rho\,\mathrm{d}z'\\ &=\frac{2\pi}{3}\int_{-2}^2\left[z'\sqrt{\rho^2+(z'-3)^2}\right]_0^{\sqrt{4-z'^2}}\,\mathrm{d}z'\\ &=\frac{2\pi}{3}\int_{-2}^2\left[z'\sqrt{13-6z'} - z'(3-z')\right]\,\mathrm{d}z'\\ &=\frac{2\pi}{3}\left[\frac{z'^3}3 - \frac32 z'^2 -\frac1{135}(13-6z')^{3/2}(9z+13)\right]_{-2}^2\\ &=\frac{2\pi}{3}\cdot\frac{64}{135}. \end{align*}

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  • $\begingroup$ I just got a solution sheet of that. it put $x' = \frac{x+2y+2z}{3}$ $y' = \frac{(2x+y-2z)}{3}$ $z' = \frac{2x-2y+z}{3}$ But I don't know the reason why it put $x', y',$ and$z '$ like the above. Does it related with the rotation? Plus Could you give me some reason that why the matrix form you offered. $\endgroup$ – se-hyuck yang Jun 5 '19 at 7:14
  • $\begingroup$ The point is precisely to put $(1,2,2)$ on a coordinate axis (I chose $z'$) to simplify the denominator and allow for some cancellation. Because the $x',y'$ in the numerator would cancel by symmetry, you have complete freedom in the other two orthonormal vectors. $\endgroup$ – user10354138 Jun 5 '19 at 7:22

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