Find volume of the body $V = \{ z = \sqrt{6-x^2-y^2}, z = x^2 + y^2 \}$ 
Find volume of the body $V = \{  z = \sqrt{6-x^2-y^2}, z = x^2 + y^2 \}$

Now what I said is:
$$V = \iint_{D} {\sqrt{6-x^2-y^2} - x^2 - y^2 dxdy}$$.
But when I wanted to get what $D$ is, I intersected the two $z$ functions to get $\sqrt{6-x^2-y^2} = x^2+y^2$.
But what is this $D$? It is certainly not a circle.

An idea: the intersection of
$$z=x^2+y^2\;,\;\;z=\sqrt{6-x^2-y^2}\implies z=\sqrt{6-z}\implies 0=z^2+z-6=(z+3)(z-2)$$
Since $\,z=-3\,$ is absurd (why?),  we then have $\;2=z=x^2+y^2\;$ and you can now get easily your limits in the $\,xy-plane\;$ by means of this canonical circle.
