Let a_r, a_ϕ, and a_z be unit vectors along r, ϕ and z directions, respectively in the cylindrical coordinate system. For the electric flux density given by D = (a_r 15 + a_ϕ 2r - a_z 3rz) Coulomb/m2, the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base at the origin is:

The correct option is 180π.

To find the total electric flux emanating from the volume enclosed by the solid cylinder, we can apply Gauss's Law, which states that the total outward electric flux is equal to the integral of the divergence of the electric flux density $D$ over the enclosed volume:

$$\Psi ={\oint }_{S}D·dS={\iiint }_V(\nabla ·D)dV$$

Given the electric flux density:

$$D=15a_r+2r{a}_{\varphi }-3rz{a}_z$$

We find the divergence of $D$ in cylindrical coordinates using the divergence formula:

$$\nabla ·D=\frac{1}{r}\frac{\partial }{\partial r}\left(r{D}_r\right)+\frac{1}{r}\frac{\partial D_{\varphi }}{\partial \varphi }+\frac{\partial D_z}{\partial z}$$

Substituting the components $D_r=15$, $D_{\varphi }=2r$, and $D_z=-3rz$:

$$\nabla ·D=\frac{1}{r}\frac{\partial }{\partial r}\left(15r\right)+\frac{1}{r}\frac{\partial }{\partial \varphi }\left(2r\right)+\frac{\partial }{\partial z}(-3rz)$$

Evaluating the partial derivatives:

$$\nabla ·D=\frac{1}{r}\left(15\right)+0-3r=\frac{15}{r}-3r$$

Now, we integrate the divergence over the volume of the solid cylinder of radius $r=3$ m and height $z=5$ m:

$$\Psi ={\int }_{z=0}^5{\int }_{\varphi =0}^{2\pi }{\int }_{r=0}^3(\frac{15}{r}-3r)rdrd\varphi dz$$

Simplifying the integrand:

$$\Psi ={\int }_0^{5}dz{\int }_0^{2\pi }d\varphi {\int }_0^3(15-3r^2)dr$$

Evaluating each integral step-by-step:

$${\int }_0^{5}dz=5$$
$${\int }_0^{2\pi }d\varphi =2\pi$$
$${\int }_0^3(15-3r^2)dr={[15r-r^3]}_0^3=\left(15\right(3)-3^3)-0=45-27=18$$

Multiplying these results together to find the total flux:

$$\Psi =5·2\pi ·18=180\pi \text{Coulomb}$$