Triple Integral Setting up bounds for region over a cone and sphere

Question

I am calculating the triple integral:

$$ \iiint\limits_{D} (z^2 + z)\ dV $$

Where D is $x^2+y^2+z^2≤4$ and $z^2≤x^2+y^2$

I understand that the first part D is the space inside or on the sphere with radius 2. For the second part of D which is a cone I am confused by how the inequality works in that sense which or side of the cone that is also satisfies the first part of D that region represents. Which is what I would like to get explained.

Answer 1

$z^2 = x^2 + y^2$ makes a double cone.

Here is a picture

Now $z^2 \le x^2 + y^2$ Would say that we are including values of $z$ such that $|z|$ is less than the radius of the cone.

Is that "inside the cone" or "outside the cone"?

Answer 2

$$\int\limits_{r=0}^2 \int\limits_{\phi = -\pi/4}^{\pi/4}\ \int\limits_{\theta = 0}^{2 \pi}\ r \left( (r \cos \phi)^2 + (r \cos \phi) \right) dr\ d \phi\ d \theta = 2 \pi \left(2+2 \sqrt{2}+\pi \right)$$