<s>[INST] Evaluating $\iiint_R\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz$ given the region $R$ 
Evaluate $$\iiint_R\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz$$, where $R$ is the region bounded by $z=x^2+y^2$ and $z=1-x^2-y^2$.

Here is what I did:  
Let: $x=r\cos\theta$, $y=r\sin\theta$, $z=z$.  
First, consider $z=x^2+y^2$.  
$r\in(0,\frac{\sqrt2}{2})$, $\theta\in(0,2\pi)$, $z\in(r^2,\frac{1}{2})$  
$$\iiint\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz=\int_{0}^{\frac{\sqrt2}{2}}\int_{0}^{2\pi}\int_{r^2}^{\frac{1}{2}}r(1-\cos\theta)\,dz\,d\theta\, dr=\frac{\pi}{8}$$
Then consider $z=1-x^2-y^2$
For $r\in(0,\frac{\sqrt2}{2})$, $\theta\in(0,2\pi)$, $z\in(1-r^2,1)$  $$\iiint\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz=\int_{0}^{\frac{\sqrt2}{2}}\int_{0}^{2\pi}\int_{1-r^2}^{1}r(1-\cos\theta)\,dz\,d\theta\, dr=\frac{\pi}{8}$$
For the final answer I added those up and I got $\frac{\pi}{8}+\frac{\pi}{8}=\frac{\pi}{4}$
Where did I do wrong?
 [/INST] Your subsitution is fine, cylindrical coordinates. But you messed up with the limits. Obviously $\theta \in (0,2\pi)$ . We have that z goes from the bottom surface to the top one, so you have to first identify them.  In this case $z=r^2$ is the one below and $z=1-r^2$ the one above, so $z \in (r^2,1-r^2)$. Then we have to define the limits for the radius. The radius goes from $0$ to the interception between both surfaces:
$$ r^2=1-r^2 \Rightarrow r=\frac{\sqrt{2}}{2} \text{     ,because  }r>0  $$ 
Considering the jacobian is $r$, the integral becomes:
$$ \int_0^{2\pi} \int_0^\frac{\sqrt{2}}{2} \int_{r^2}^{1-r^2} (1-\cos{\theta})\space r \space dz\space dr \space d\theta $$
The cosine part is $0$ because of the symmetry, and then we have that:
$$ \text{Integral} =2\pi \int_0^{\frac{\sqrt{2}}{2}} (1-2r^2) \space rdr = \frac{\pi}{2}$$
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