Answer
$$8\pi $$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^2_0 \int^{\sqrt{4-r^2}}_{r-2} \int^{2\pi}_0 (r (\sin\theta)+1) \space r \space d\theta \space dz \space dr =\int^2_0 \int^{\sqrt{4-r^2}}_{r-2} 2 \pi \times r \space dz \space dr \\= 2\pi \int^2_0 [r(4-r^2)^{1/2}-r^2+2r] \space dr \\=2\pi[-\dfrac{1}{3} \times (4-r^2)^{3/2}-\dfrac{1}{3} \times r^3+r^2]^2_0 \\=2\pi[-\dfrac{8}{3}+4+\dfrac{1}{3}(4)^{3/2}] \\=8\pi $$
