Question

Evaluate the triple integral:
Integral from (0 to pi/2) integral from (0 to 1) integral from (0 to sqrt(1-r^2)) of r * sqrt(r^2+z^2) dz dr dtheta

Answer 1

Just formatting it for you :)
easier to read this way.
\[\Large\rm \int\limits_0^{\pi/2}\int\limits_0^1\int\limits_0^{\sqrt{1-r^2}}~r\sqrt{r^2+z^2}~dz~dr~d\theta\]

Answer 2

@zepdrix thank you :)

Answer 3

@zepdrix help him, please!! I don't know it. :)

Answer 4

Hmm our innermost integral seems a lil tough since we don't have a z outside the root :(
\[\Large\rm \int\limits_0^{\sqrt{1+r^2}} r\sqrt{r^2+z^2}dz\]I guess we'll have to apply a trig substitution.\[\Large\rm z=r \tan \phi\]\[\Large\rm dz=r \sec^2 \phi ~d \phi\]Changing our integral to,\[\Large\rm \int\limits r^3 \sec^3\phi ~d\phi\]Something like that? :d
Maybe there's an easier approach to this, I'm not seeing it though.

Answer 5

@OOOPS thank you soooo much :) I've been stuck on this problem for a really long time