Think of analagous situation where you have int(z)

Where R is bounded by the paraboloid z=4y^2+4x^2 and the plane z=4.

Then it is fairly obvious to see

z varies from 4y^2 + 4x^2 to 4

the outer double integral is then over the circle of radius 2 in the xyplane

converting to cylindrical int(zr dzdrd(theta)

z varies from 4r^2 to 4

r from 0 to 1

theta from 0 to 2pi

The only differnce in your problem is that the parabaloid comes out in the x direction and you end up integrating over a circle of radius 2 in the y -z plane --interchange the roles of x and z

You have

int(xr dxdrd(theta)

x varies from 4r^2 to 4

r from 0 to 1

theta from 0 to 2pi

To answer your second question it depends on whether you have a triple

or double integral with a function of z in the integrand.

If you have a triple integral such as in the analagous problem I suggested

then you are using cylindrical coordinates where z = z and you just integrate it out

If you have a double integral then you will have a surface or surfaces

where z is a known function of x and y such as z = x^2 + y^2 = r^2

in which case you convert z to functions of x and y (or r if you're going polar)

Eg dblint(zrdrdtheta) would be dblint(r^2 rdrdtheta) = dblint(r^3dr dtheta)

Hope this helps