Suppose $T$ is a region bounded by $ z = x^2 + y^2 $ and $z = 1 - x^2 - y^2$.
I know they intersect at $2x^2 + 2y^2 = 1 $ which is a circle of radius $1/4$. Also, we see $0 \leq z \leq 1 $. Also, we have
$$ - \sqrt{ 1/2 - x^2 } \leq y \leq \sqrt{1/2 - x^2 } $$ $$ - \sqrt{ 1/2 - y^2} \leq x \leq \sqrt{1/2 - y^2} $$
are these correct bounds for my variables $x,y,z$? thanks
Added: My goal is to compute
$$ \int \int \int_T \frac{x}{x^2+y^2}dxdydz $$
So I need to find the bounds for $T$. But,with this bounds that I found the integral looks very hard to compute.
Approaching the problem like this won't get you far, so the question of your integration bounds being correct or not is somehow irrelevant. It is better to resort to a change of variables and perform integration in cylindrical variables. In order to do this, let $x = r \cos u$ and $y = r \sin u$, with $z$ staying unchanged, the domain for $(r, u)$ being $(0, \infty) \times (0, 2 \pi)$. The two surfaces (they are paraboloids) that delimit your domain of integration become $z = r^2$ and $z = 1 - r^2$. Notice that not only are they easier to write (and work with), but you also have the advantage of getting rid of $u$ (because the domain of integration is rotationally invariant around the $z$ axis).
The two surfaces intersect when $r^2 = 1 - r^2$, which means $r = \frac 1 {\sqrt 2}$ (because $r$ must be positive), so this means that $0 < r \le \frac 1 {\sqrt 2}$. For each such $r$, $z$ varies between $r^2$ and $1-r^2$. This means that your integral becomes
$$\int \limits _0 ^{2\pi} \int \limits _0 ^{\frac 1 {\sqrt 2}} \int \limits _{r^2} ^{1-r^2} \frac {r \cos u} {r^2} r \ \Bbb d z \ \Bbb d r \ \Bbb d u ,$$
where the $r$ after the fraction is the modulus of the Jacobian of the change of variables. Performing the integral in $u$ first gives us the result $0$, because $\int _0 ^{2\pi} \cos u \ \Bbb d u = \sin u \big| _0 ^{2\pi} = 0$, so it becomes unnecessary to compute the remaining two integrals. The final result will be $0$.