# How to solve this triple intergral $I=\iiint_{Q}\frac{1}{x^2+y^2}dv$?

$$I \equiv \iiint_{Q}\frac{{\rm d}v}{x^2+y^2}$$ 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.

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