Let $A=\{(x,y,z) \space\space s.t \space\space x^2+y^2<1 \space\space and \space\space 0<z<1\}$, and $f(x,y,z)=z$. I want to determine $\int_A f$ with integration by projection. More precisely, I have the projection of $A$ on to the $xy$-plane, the set $$A(x)=\{(x,y)\in \Bbb{R^2}\vert(x,y,z)\in A\}=B(0,1).$$
The projection of $A$ onto the $z$-axis is $$P_1 = \{z\in \Bbb{R}\vert\exists(x,y) \space\space s.t \space\space (x,y,z)\in A\}.$$
Then I the integral by projection should be $$\int_A f=\int_{P_1}\Biggr(\int_{A(x)}f(x,y)dy\Biggr)dx.$$
I dont know how to develop the integral from here. The projections are used to establish bounds for integration. The theorem I am using requires that $f(x,y,z)\in \cal{R(A)}$, and $f(x,y) \in \cal{R(A(x))}.$ I don't see why we need to establish this kind of function. Do we need a different function with respect to each projection?
I have never seen the phrase "integration by projection" before. The region of integration is, of course, that part of the cylinder, with axis along the z-axis and radius 1, between z= 0 and z= 1. In this particular case, the function to be integrated is simply f(x,y,z)= z so we could write this as $\int_{x=-1}^1\int_{y= -\sqrt{1- x^2}}^{\sqrt{1- x^2}} \int_{z= 0}^1 z dzdydx$. You could also do the integration in x and y in polar coordinates as $\int_{r= 0}^{1}\int_{\theta= 0}^{2\pi}\int_{z= 0}^1 z dzd\theta dr$. Or you could just write the region in the xy-plane as "R" and call it $\int_R \int_0^1 z dzdA$ where dA is the differential in the xy-plane.
Of course that is the same as $\left(\int_R dA\right)\left(\int_0^1 z dz\right)$. Of course, $\int_R dA$ is just the area of the cirle with radius 1, $2\pi$ and $\int_0^1 z dz= \left[\frac{1}{2}z^2\right]_0^1= \frac{1}{2}$.
The value of your integral is $(2\pi)\left(\frac{1}{2}\right)= \pi$.