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Good night, i have a problem solving this integral:

$$\int_{1}^{2}\int_{x}^{2x}\int_{\sqrt{1-x^{2}-y^{2}}}^{\sqrt{2xy}}\frac{zdzdydx}{x^{2}+y^{2}+z^{2}}$$

I think make a change to spherical coordinate but, I don't know how I can calculate the integration limits. Please, help me!

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First, let $u=\rho^2=x^2+y^2+z^2$, then $du/dz=2z$ so $zdz=du/2$. So the inner $dz$ integral is $$ \int (1/u) du/2 = (1/2)\ln u = \ln (\rho^2)/2 = \ln \rho, $$ and the definite version of the inner integral is $$ [\ln\rho]_{\rho=1}^{\rho^2=x^2+y^2+\sqrt{2xy}^2} = [\ln\rho]_{\rho=1}^{\rho^2=x^2+y^2+2xy} $$ $$ =[\ln\rho]^{\rho^2=(x+y)^2}_{\rho=1} = \ln(x+y). $$ So you're left with $$ \int_1^2\int_x^{2x} \ln(x+y)\,dy\,dx.$$

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  • $\begingroup$ Excellent Man! very good, but... i don't understand this: $\left[ln\rho\right]\mid_{\rho=1}^{\rho=x^{2}+y^{2}+\sqrt{2xy}} $ why?? $\endgroup$ – Bvss12 May 25 '16 at 1:09
  • $\begingroup$ you're right, i've fixed it now $\endgroup$ – Bjørn Kjos-Hanssen May 25 '16 at 1:12
  • $\begingroup$ Why you erase the comment? Please, rewrite ): $\endgroup$ – Bvss12 May 25 '16 at 1:15
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    $\begingroup$ $z=\sqrt{2xy}$ means $x^2+y^2+z^2=x^2+y^2+\sqrt{2xy}^2$, right? And $z=\sqrt{1-x^2-y^2}$ means $x^2+y^2+z^2=1$. That's all. $\endgroup$ – Bjørn Kjos-Hanssen May 25 '16 at 1:17
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    $\begingroup$ oh man! thanks you're a genious :D $\endgroup$ – Bvss12 May 25 '16 at 1:18

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