The formula is: $$ \iiint_G {|xy| \over{x^2}} dxdydz $$ $G$ is defined by the inequality: $\sqrt{x^2+y^2} < z < \sqrt{1-x^2-y^2}$
I tried to use this substitute: $$ \begin{cases} x = rsin(\phi)\\ y = rcos(\phi)\\ z = z \end{cases} $$
So $G$ transforms to $r < z < \sqrt{1 - r^2}$, but for me it's not much better. Other suggestions?
Thanks for any help!
$$G_1=\left\{(\theta,r,z)\Big{|}\,\,\frac{\pi}{4}\le\theta\le\frac{3\pi}{4} ,\,0\le r\le 1 \, , r\le z\le\sqrt{1-r^2}\right\}\\ G_2=\left\{(\theta,r,z)\Big{|}\,\,\frac{5\pi}{4}\le\theta\le\frac{7\pi}{4} ,\,0\le r\le 1 \, , r\le z\le\sqrt{1-r^2}\right\} $$ We have $G=G_1\cup G_2$.Since $G$ is symmetric with respect $x-$ axes and $y-$ axes, thus $$ \iiint_G {|xy| \over{x^2}} dxdydz=4\int_{0}^{\frac {\pi}{4}}\int_{0}^{1} \int_{r}^{\sqrt{1-r^2}}\tan\theta \,\text{dz dr d}\theta $$