Question

Given, \( \vec{\phi} = xy\hat{i} + yz\hat{j} + xz\hat{k} \). S is a surface bounded by the planes x = 0, y = 0, z = 0, x = 3, y = 2, and z = 1. If \( \hat{n} \) is the unit vector normal to S, then \( \iint_S \vec{\phi} . \hat{n} dS \) is

• A. 18
• B. 9
• C. 36
• D. 3

Hint:

For calculating the flux through a simple closed surface like a sphere, box, or cylinder, the Divergence Theorem is almost always the easiest path. It turns a potentially complex surface integral (often requiring six separate integrals for a box) into a single, straightforward triple integral.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks to calculate the surface integral of a vector field over a closed surface S. The surface S is a rectangular box (a cuboid). This integral represents the net flux of the vector field \( \vec{\phi} \) out of the closed surface. The Gauss Divergence Theorem provides a powerful tool to solve such problems by converting the surface integral into a volume integral.
Step 2: Key Formula or Approach:
The Gauss Divergence Theorem states: \[ \iint_S \vec{\phi} . \hat{n} dS = \iiint_V (\nabla . \vec{\phi}) dV \] where V is the volume enclosed by the surface S, and \( \nabla . \vec{\phi} \) is the divergence of the vector field \( \vec{\phi} \).
The divergence is calculated as: \[ \nabla . \vec{\phi} = \frac{\partial \phi_x}{\partial x} + \frac{\partial \phi_y}{\partial y} + \frac{\partial \phi_z}{\partial z} \] Step 3: Detailed Calculation:
First, let's find the divergence of the given vector field \( \vec{\phi} = xy\hat{i} + yz\hat{j} + xz\hat{k} \). The components are \( \phi_x = xy \), \( \phi_y = yz \), and \( \phi_z = xz \). \[ \nabla . \vec{\phi} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(xz) \] \[ \nabla . \vec{\phi} = y + z + x \] Next, we apply the Divergence Theorem. We need to integrate the divergence over the volume V of the rectangular box. The box is defined by the limits:
- \( 0 \le x \le 3 \)
- \( 0 \le y \le 2 \)
- \( 0 \le z \le 1 \)
The volume integral is: \[ \iiint_V (x + y + z) dV = \int_{z=0}^{1} \int_{y=0}^{2} \int_{x=0}^{3} (x + y + z) \,dx \,dy \,dz \] We evaluate the integral step-by-step, starting from the inside:
1. Integrate with respect to x:
\[ \int_{0}^{3} (x + y + z) \,dx = \left[ \frac{x^2}{2} + yx + zx \right]_{0}^{3} = \left( \frac{3^2}{2} + 3y + 3z \right) - 0 = \frac{9}{2} + 3y + 3z \] 2. Integrate with respect to y:
\[ \int_{0}^{2} \left( \frac{9}{2} + 3y + 3z \right) \,dy = \left[ \frac{9}{2}y + \frac{3y^2}{2} + 3zy \right]_{0}^{2} \] \[ = \left( \frac{9}{2}(2) + \frac{3(2^2)}{2} + 3z(2) \right) - 0 = 9 + \frac{12}{2} + 6z = 9 + 6 + 6z = 15 + 6z \] 3. Integrate with respect to z: \[ \int_{0}^{1} (15 + 6z) \,dz = \left[ 15z + \frac{6z^2}{2} \right]_{0}^{1} = \left[ 15z + 3z^2 \right]_{0}^{1} \] \[ = (15(1) + 3(1)^2) - 0 = 15 + 3 = 18 \] Step 5: Why This is Correct:
The application of the Gauss Divergence Theorem is the correct and most efficient method. The calculation for the given vector field yields 18.