$$Integral = \iiint_V x+z\;dV $$
$$Region$$ $$x^2 + y^2 + z^2 \leq 1$$
$$z \leq \sqrt{x^2 + y^2}$$
Geogebra shows that the region looks like in the picture below but given the inquality sign in the equations, is the actual region inside the cone and sphere (icecream cone) or is the actual region inside the sphere but minus the icecream cone? Initially I thought it is the first one but given the inquality, it made me think it is the second one. Which one is it? Thank you very much!
For completeness, here's an answer based on my comment above.
Because of the inequality $x^2 + y^2 + z^2 \leq 1$, we know the region is contained inside the unit sphere. We also know that $z = \sqrt{x^2 + y^2}$ describes the surface of the cone in the above picture. So the question is: is the region equal to the ice cream cone (i.e. above the cone surface) or is it equal to the sphere minus the ice cream cone (i.e. below the cone surface).
An easy way to determine the answer is to take a point in one of these two regions and check whether it satisfies the inequalities. For example, if we take a point with $x=y=0$ and $0<z<1$, we can see that it lies inside the ice cream cone. However, we clearly have $$z > \sqrt{x^2 + y^2} = 0.$$ In other words, this point does not satisfy the inequality. Therefore, we can conclude that the region of integration is equal to the sphere minus the ice cream cone.