Double integral - Volume

Question

I find it very difficult to solve this problem. I need help setting the integral up. I know that the first one is cone and the third one is paraboloid, but I can't define the limits. Some explanation would be nice, too. Thanks in advance.

Find the volume of the solid between the surfaces, using double integral

$z^2=x^2+y^2,\\z\geq 0\\2-z=x^2+y^2$

Answer 1

HINT: Set this up in cylindrical coordinates. The two surfaces intersect when $z^2=2-z$, so you can figure out the limits on $r$ from this.

Answer 2

Can you see what I've done so far? Is this the right way?

$z^2=x^2+y^2,\\z\geq 0\\2-z=x^2+y^2$

$z^2+z-2=0\\z=1 => x^2+y^2 = 1$

$x=rcosa\\y=rsina\\|Y|=r$

$0\le r\le 1\\0\le a\le2\pi$

$V=\int\int f1(x,y)-f2(x,y)dxdy$

$\sqrt{x^2+y^2}=r\\2-x^2-y^2 = 2-r^2=> V=\int_0^{2\pi}da\int_0^1 (2-r^2-r)dr$