Answer
$$\dfrac{5\pi}{2}$$
Work Step by Step
To intersect the paraboloids, we have
$4x^2+4y^2=5-x^2-y^2 \implies x^2+y^2=1$
or, $ z=4$
Our aim is to integrate the integral as follows:
$$ Volume =4\int^{\pi/2}_0 \int^1_0 \int^{5-r^2}_{4r^2}dz \space r \space dr \space d\theta \\=(4) \times \int^{\pi/2}_0 \int^1_0 (5r-5r^3)dr \space d\theta \\=(20) \times \int^{\pi/2}_0 [\dfrac{r^2}{2}-\dfrac{r^4}{4} \space d\theta] \\=(5) \int^{\pi/2}_0 (1) d\theta \\=\dfrac{5\pi}{2}$$
