12.3#15

I don't understand this problem. I've tried multiplying the base and hieght and tried to do this:
I (pi 0) I (8 0 )(x=rcosA - r^2)r dr dz dA

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well, its hard to figure out what you mean but I guess maybe you mean A=θ, and I guess you're doing a triple integral, which could work but is excessive for a homework problem dealing with polar integrals (and wrong as you've implemented here).

From the way the plane z=x cuts the xy-plane at the y-axis,  and from the fact that the region is above the xy-plane, you have a base that's a semidisk in the x≥0 part of the xy-plane and a height (z) that's equal to x above that, which gives the integrals

∫_{-π/2}^{π/2} ∫_0^8 r cos(θ) r dr dθ,

so you kind of have the integrand except for that -r^2) part.  The limits of integration in r are fine, but the limits of integration in θ would be for the plane z=y instead of z=x.