Answer:
Q=185.84C
Explanation:
We have to take into account the integral
[tex]Q=\int \rho dV[/tex]
In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4
[tex]Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx[/tex]
but, for symmetry:
[tex]Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C[/tex]
HOPE THIS HELPS!!
