evaluate the following integral in cylindrical coordinates triple integral 1/(1+x^2+y^2) dzdxdy z=0..2 y=0..square root(1-x^2) x=-1..1

using pi as needed

Respuesta :

In cylindrical coordinates, we take

[tex]x=r\cos\theta[/tex]

[tex]y=r\sin\theta[/tex]

[tex]z=z[/tex]

so that [tex]\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz[/tex].

We have

[tex]1+x^2+y^2=1+r^2[/tex]

and the integral is

[tex]\displaystyle\int_0^2\int_0^\pi\int_0^1\frac r{1+r^2}\,\mathrm dr\,\mathrm d\theta\,\mathrm dz=\frac{\ln2}2\int_0^2\int_0^\pi\mathrm d\theta\,\mathrm dz=\boxed{\pi\ln2}[/tex]