Answer
$104$
Work Step by Step
$$I_0= \int_0^{2\pi} \int_{2x}^{4} (x^2+y^2) (3) \ dy \ dx \\= 3 \int_0^{2} (4x^2 +\dfrac{64}{3}-\dfrac{14x^3}{3}) \ dx \\=3 [\dfrac{4}{3}x^3+\dfrac{64}{3}\times x^3-\dfrac{14}{12} \times x^4]
_0^2 \\=3 [\dfrac{4(8-0)}{3}+\dfrac{64(2-0)}{3}-\dfrac{14(16-0)}{12}] \\=104$$
