Answer
$$12\pi $$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^1_{-1} \int^\sqrt{z}_0 \int^{1+ \cos\theta}_0 (4r) \space dr \space d \theta\space dz = (2) \times \int^1_{-1} \int^{2\pi}_0 (1+\cos\theta)^2 \space d\theta \space dz \\= (6) \times \int^1_{-1}\pi \space d \theta \\=12\pi $$
