$x^2+y^2\le 1$; $z=\sqrt{x^2+y^2}$; and $x^2+y^2=4-z$

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I need to find a value and "surface" of a body which is contained in the following contours: $x^2+y^2\le 1$; $z=\sqrt{x^2+y^2}$; and $x^2+y^2=4-z$. Some hints and directions will be helpful. Sorry for my English.

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The surface $x=\sqrt{x^2+y^2}$ is the "bottom". The surface $x^2+y^2=1$ is the "side". And $z=4-x^2-y^2$ is the "ceiling".

So the volume is $$ V=\int_0^{2\pi}\int_0^1[(4-r^2)-r]\,r\,dr. $$ For the surface, you will have to find the area of each of the bottom, side, and ceiling. For this, you will need to calculate the intersections.