Question

A vector field \[ \mathbf{B}(x, y, z) = x \mathbf{\hat{i}} + y \mathbf{\hat{j}} - 2z \mathbf{\hat{k}} \] 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 \(\mathbf{n}\) denotes the unit outward normal to the curved surface S of the cone, the value of the integral \[ \iint_S \mathbf{B} \cdot \mathbf{n} \, dS \] equals ................ (Answer in integer) 

Hint:

When asked to compute a flux integral over an open surface, always check if the Divergence Theorem can simplify the problem. Calculate the divergence \(\nabla \cdot \mathbf{F}\). If it's zero or a simple constant, using the theorem by closing the surface is almost always easier than parameterizing the original surface.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the flux of a vector field \(\mathbf{B}\) through an open surface \(S\), which is the curved lateral surface of a cone. A direct calculation of this surface integral would be complex. A more straightforward approach is to use the Gauss Divergence Theorem. The theorem relates the flux through a closed surface to the divergence of the field within the enclosed volume.
Step 2: Key Formula or Approach:
1. Gauss Divergence Theorem: For a closed surface \(S_{closed}\) enclosing a volume \(V\), the outward flux is given by: \[ \oiint_{S_{closed}} \mathbf{B} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{B}) \, dV \] 2. Closing the Surface: Our surface \(S\) is open. We can create a closed surface by adding the flat circular base of the cone, let's call it \(D\). The closed surface is then \(S_{closed} = S \cup D\).
3. Applying the Theorem: The total flux through the closed surface is the sum of the flux through the curved part and the flux through the base: \[ \oiint_{S \cup D} \mathbf{B} \cdot \mathbf{n} \, dS = \iint_S \mathbf{B} \cdot \mathbf{n} \, dS + \iint_D \mathbf{B} \cdot \mathbf{n} \, dS \] 4. By rearranging, we get the integral we want: \[ \iint_S \mathbf{B} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{B}) \, dV - \iint_D \mathbf{B} \cdot \mathbf{n} \, dS \] Step 3: Detailed Calculation:
Part 1: Calculate the Divergence
The vector field is \(\mathbf{B} = x\mathbf{i} + y\mathbf{j} - 2z\mathbf{k}\). The divergence is: \[ \nabla \cdot \mathbf{B} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(-2z) = 1 + 1 - 2 = 0 \] Part 2: Calculate the Volume Integral
Since the divergence is zero everywhere, the volume integral is zero: \[ \iiint_V (\nabla \cdot \mathbf{B}) \, dV = \iiint_V 0 \, dV = 0 \] Part 3: Relate the Integrals
From the Divergence Theorem, we now have: \[ \iint_S \mathbf{B} \cdot \mathbf{n} \, dS + \iint_D \mathbf{B} \cdot \mathbf{n} \, dS = 0 \] \[ \implies \iint_S \mathbf{B} \cdot \mathbf{n} \, dS = - \iint_D \mathbf{B} \cdot \mathbf{n} \, dS \] We just need to calculate the flux through the base disk \(D\). Part 4: Calculate the Flux through the Base Disk D
- The base \(D\) is a circle in the \(xy\)-plane, so \(z=0\) for all points on \(D\).
- The outward unit normal vector to the volume through the base points in the negative z-direction. So, \(\mathbf{n} = -\mathbf{k}\).
- Evaluate the vector field \(\mathbf{B}\) on the surface \(D\) (where \(z=0\)): \[ \mathbf{B}|_{z=0} = x\mathbf{i} + y\mathbf{j} - 2(0)\mathbf{k} = x\mathbf{i} + y\mathbf{j} \] - Calculate the dot product \(\mathbf{B} \cdot \mathbf{n}\): \[ \mathbf{B} \cdot \mathbf{n} = (x\mathbf{i} + y\mathbf{j}) \cdot (-\mathbf{k}) = 0 \] - The flux through the base is therefore zero: \[ \iint_D \mathbf{B} \cdot \mathbf{n} \, dS = \iint_D 0 \, dS = 0 \] Part 5: Find the Final Value
Substitute the result from Part 4 into the equation from Part 3: \[ \iint_S \mathbf{B} \cdot \mathbf{n} \, dS = - (0) = 0 \] Step 4: Final Answer:
The value of the integral is 0.
Step 5: Why This is Correct:
The divergence of the vector field is zero, which means the total flux out of any closed volume is zero (the field is "solenoidal"). We also showed that the flux through the base of the cone is zero. Therefore, the flux through the remaining curved surface must also be zero to maintain a total flux of zero.