A vector field

𝐁(𝑥, 𝑦, 𝑧) = 𝑥 𝑖̂+ 𝑦 ĵ− 2𝑧 k̂

is defined over a conical region having height ℎ = 2, base radius 𝑟 = 3 and axis along z, as shown in the figure. The base of the cone lies in the x-y plane and is centered at the origin.

If 𝒏 denotes the unit outward normal to the curved surface 𝑆 of the cone, the value of the integral

                               ∫_s𝐁 S ⋅ 𝒏 dS    equals _________ . (Answer in integer)

The correct answer is 0.

Step-by-step Explanation:

1. Understanding the Geometry and Vector Field from the Diagram
As shown in the first image, we have a conical region with height $h=2$, base radius $r=3$, and its axis along the z-axis. The base of the cone, centered at the origin $O$, lies in the x-y plane (where $z=0$). The curved surface is denoted by $S$.
The given vector field is:
$$\mathbf{B}(x,y,z)=x\hat{i}+y\hat{j}-2z\hat{k}$$

2. Applying the Gauss Divergence Theorem
To find the flux through the curved surface $S$, we can define a closed surface bounding the conical volume $V$. The total boundary of the volume consists of two parts:
- The curved surface $S$
- The flat circular base $S_{\text{base}}$ at the bottom ($z=0$)

By the Gauss Divergence Theorem, the net outward flux of the vector field $\mathbf{B}$ through the entire closed surface is equal to the volume integral of the divergence of $\mathbf{B}$ over the enclosed volume $V$: $${\iint }_S\mathbf{B}\cdot \mathbf{n}dS+{\iint }_{S_{\text{base}}}\mathbf{B}\cdot {\mathbf{n}}_{\text{base}}dS={\iiint }_V(\nabla \cdot \mathbf{B})dV$$

3. Calculating the Divergence of the Vector Field
Let us find the divergence of $\mathbf{B}$: $$\nabla \cdot \mathbf{B}=\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(y\right)+\frac{\partial }{\partial z}(-2z)$$
Evaluating the partial derivatives: $$\nabla \cdot \mathbf{B}=1+1-2=0$$
Since the divergence is zero everywhere, the volume integral is: $${\iiint }_V(\nabla \cdot \mathbf{B})dV=0$$

4. Evaluating Flux through the Circular Base
The base $S_{\text{base}}$ lies in the x-y plane where $z=0$. The unit outward normal vector pointing away from the conical volume at the base is directed in the negative z-direction:
$${\mathbf{n}}_{\text{base}}=-\hat{k}$$
Taking the dot product of the vector field $\mathbf{B}$ with ${\mathbf{n}}_{\text{base}}$ gives: $$\mathbf{B}\cdot {\mathbf{n}}_{\text{base}}=(x\hat{i}+y\hat{j}-2z\hat{k})\cdot (-\hat{k})=2z$$
Evaluating this expression at the base where $z=0$ yields: $$\mathbf{B}\cdot {\mathbf{n}}_{\text{base}}=2\left(0\right)=0$$
Therefore, the flux through the base of the cone is: $${\iint }_{S_{\text{base}}}\mathbf{B}\cdot {\mathbf{n}}_{\text{base}}dS=0$$

5. Computing Flux through the Curved Surface
Substituting these values back into the Gauss Divergence Theorem equation: $${\iint }_S\mathbf{B}\cdot \mathbf{n}dS+0=0$$
Which simplifies to: $${\iint }_S\mathbf{B}\cdot \mathbf{n}dS=0$$