Set Up the Three-Dimensional Integral $\int _ { R } y d V$

Set up the three-dimensional integral $\int _ { R } y d V$ where R is the "ice-cream cone" enclosed by a sphere of radius 2 centered at the origin and the cone $z = \sqrt { 3 x ^ { 2 } + 3 y ^ { 2 } }$ .Use rectangular coordinates.

A) $\int_{R} y d \mathrm{~V}=\int_{0}^{2 \pi} \int_{0}^{1} \int_{\sqrt{3 r}}^{\sqrt{4-r^{2}}} r^{2} \sin \theta d z d r d \theta$
B) $\int _ { R } y d V = \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \pi / 6 } \int _ { 0 } ^ { 2 } \rho ^ { 3 } \sin \theta \sin ^ { 2 } \phi d \rho d \phi d \theta$
C) $\int _ { R } y d V = \int _ { - 1 } ^ { 1 } \int _ { - \sqrt { 1 - x ^ { 2 } } } ^ { \sqrt { 1 - x ^ { 2 } } } \int _ { \sqrt { 3 x ^ { 2 } + 3 y ^ { 2 } } } ^ { \sqrt { 4 - x ^ { 2 } - y ^ { 2 } } } y d z d y d x$