2
$\begingroup$

$$ I \equiv \iiint_{Q}\frac{{\rm d}v}{x^2+y^2} $$ Which $Q$ is a solid bounded above by $z=4-x^2-y^2$ and below by the sphere $x^2+y^2+z^2=9$.

I have tried with this multiple integration by using cylindrical coordinate. As a result, I got

$$ I=\int_{0}^{2\pi}\int \int_{\sqrt{9-r^{2}}}^{4-r^{2}}\,\,\frac{1}{r} \,{\rm d}z\,{\rm d}r\,{\rm d}\theta $$

I cannot determine the value of $r$ and I cannot solve this integral. Please give me some hints or other way to solve this triple integral. Thank in advanced.

I have written with wrong sign. It was $z=4-x^2+y^2$ Here is right problem. Actually, $Q$ is a solid bounded above by $z=4-x^2+y^2$ and bounded below the sphere $x^2+y^2+z^2=9$.

The red dot that I noted in this graph. It is a region that I intended. Region picture

$\endgroup$
8
  • $\begingroup$ You could also try with spherical coordinates if cilindrical doesn't work out. $\endgroup$ – Measure me Jan 16 at 15:28
  • 2
    $\begingroup$ The integral doesn't converge $\endgroup$ – Raffaele Jan 16 at 16:31
  • $\begingroup$ Is $Q$ all the points below the surface and outside the sphere, or are there other constraints e.g. $z\ge0$ or $r\le4$? $\endgroup$ – J.G. Jan 16 at 17:19
  • $\begingroup$ @J.G. math3d.org/ZwTf1nHV For my intended, I think it bounded between sphere and surface. $\endgroup$ – Matin Jan 16 at 17:30
  • $\begingroup$ @J.G. $Q$ is all the points above the surface and inside the sphere. $\endgroup$ – Matin Jan 16 at 17:32

Your Answer

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

Browse other questions tagged or ask your own question.