Good night, i have a problem solving this integral:


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


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 $$ \int_1^2\int_x^{2x} \ln(x+y)\,dy\,dx.$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ $z=\sqrt{2xy}$ 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
  • 1
    $\begingroup$ oh man! thanks you're a genious :D $\endgroup$ – Bvss12 May 25 '16 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.