Let aᵣ, a${}_{\text{ϕ},}$ 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 = (aᵣ 15 + a${}_{\text{ϕ}}$ 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:

To determine the total electric flux emanating from the given volume, we can use Gauss's Law in its differential form, also known as the Divergence Theorem. This theorem states that the total outward electric flux of an electric flux density field D through a closed surface is equal to the volume integral of the divergence of D over the volume enclosed by that surface.

The formula for the total electric flux ${\Phi }_E$ is:

$${\Phi }_E={\oint }_S\mathbf{D}\cdot d\mathbf{S}={\int }_V(\nabla \cdot \mathbf{D})dV$$

Here, $\mathbf{D}$ is the electric flux density, $V$ is the volume of the cylinder, and $\nabla \cdot \mathbf{D}$ is the divergence of $\mathbf{D}$.

Electric Flux Density Components

The given electric flux density D is provided in the cylindrical coordinate system:

$$\mathbf{D}=(a_{r}15+a_{\phi }2r-a_{z}3rz){\text{ Coulomb/m}}^2$$

From this expression, we can identify the individual components of the electric flux density vector:

- The radial component, $D_r=15$
- The azimuthal component, $D_{\phi }=2r$
- The axial component, $D_z=-3rz$

Divergence Calculation in Cylindrical Coordinates

The divergence of a vector field D in cylindrical coordinates $(r,\phi ,z)$ is calculated using the following formula:

$$\nabla \cdot \mathbf{D}=\frac{1}{r}\frac{\partial }{\partial r}(r{D}_r)+\frac{1}{r}\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_z}{\partial z}$$

Let's calculate each term by substituting the components of D:

- First term (radial component): $\frac{1}{r}\frac{\partial }{\partial r}(r{D}_r)$ Substitute $D_r=15$: $$\frac{1}{r}\frac{\partial }{\partial r}(r\cdot 15)=\frac{1}{r}\frac{\partial }{\partial r}(15r)=\frac{1}{r}(15)=\frac{15}{r}$$
- Second term (azimuthal component): $\frac{1}{r}\frac{\partial D_{\phi }}{\partial \phi }$ Substitute $D_{\phi }=2r$: $$\frac{1}{r}\frac{\partial }{\partial \phi }(2r)=\frac{1}{r}(0)=0$$ This term is zero because $2r$ does not vary with the azimuthal angle $\phi$.
- Third term (axial component): $\frac{\partial D_z}{\partial z}$ Substitute $D_z=-3rz$: $$\frac{\partial }{\partial z}(-3rz)=-3r$$

Combining these calculated terms, the divergence of the electric flux density D is:

$$\nabla \cdot \mathbf{D}=\frac{15}{r}+0-3r=\frac{15}{r}-3r$$

Total Electric Flux Integration

The total electric flux ${\Phi }_E$ is obtained by integrating the divergence $(\nabla \cdot \mathbf{D})$ over the entire volume of the solid cylinder. The cylinder has a radius of $R=3$ m and a height of $H=5$ m, and it is oriented along the z-axis with its base at the origin.

The limits of integration for the cylindrical coordinates $(r,\phi ,z)$ are:

- Radius $r$: from $0$ to $3$
- Azimuthal angle $\phi$: from $0$ to $2\pi$
- Height $z$: from $0$ to $5$

The differential volume element in cylindrical coordinates is $dV=rdrd\phi dz$.

Setting up the volume integral for the total electric flux:

$${\Phi }_E={\int }_V(\nabla \cdot \mathbf{D})dV={\int }_{z=0}^5{\int }_{\phi =0}^{2\pi }{\int }_{r=0}^3\left(\frac{15}{r}-3r\right)rdrd\phi dz$$

First, simplify the integrand by multiplying $\left(\frac{15}{r}-3r\right)$ by $r$:

$${\Phi }_E={\int }_{z=0}^5{\int }_{\phi =0}^{2\pi }{\int }_{r=0}^3(15-3r^2)drd\phi dz$$

Step-by-step Integration

**Integration with respect to $r$:**

We integrate the innermost integral first:

$${\int }_{r=0}^3(15-3r^2)dr={\left[15r-\frac{3r^3}{3}\right]}_0^3={\left[15r-r^3\right]}_0^3$$

Now, evaluate this expression at the upper and lower limits:

$$(15\cdot 3-3^3)-(15\cdot 0-0^3)=(45-27)-(0)=18$$

**Integration with respect to $\phi$:**

Next, we integrate the result from the previous step with respect to $\phi$:

$${\int }_{\phi =0}^{2\pi }18d\phi =[18\phi {]}_0^{2\pi }$$

Evaluate at the limits:

$$18(2\pi )-18(0)=36\pi$$

**Integration with respect to $z$:**

Finally, we integrate the result with respect to $z$:

$${\int }_{z=0}^{5}36\pi dz=[36\pi z{]}_0^5$$

Evaluate at the limits:

$$36\pi (5)-36\pi (0)=180\pi$$

Therefore, the total electric flux emanating from the volume enclosed by the solid cylinder is $180\pi$ Coulomb.