Question

Let S be the surface of the cube bounded by \( x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 \). The value of the surface integral \( \iint_S \vec{F} \cdot d\vec{s} \) of a vector field \( \vec{F} = 4xz\hat{i} - y^2\hat{j} + yz\hat{k} \) over the entire surface S of the cube, is \dots\dots.

• A. \( \frac{5}{2} \)
• B. \( 1 \)
• C. \( \frac{1}{2} \)
• D. \( \frac{3}{2} \)

Hint:

Always check if the surface is closed. If it forms an enclosed volume (like a cube, sphere, or cylinder), applying the Divergence Theorem avoids computing separate flux integrals for each individual face.

Answer 1

The Correct Option is D

Concept: Evaluating a surface integral over a closed multi-faced surface like a cube can be tedious. The Divergence Theorem (Gauss's Theorem) simplifies this by relating the surface integral of a vector field over a closed surface \( S \) to a volume integral over the region \( V \) enclosed by \( S \): \[ \iint_S \vec{F} \cdot d\vec{s} = \iiint_V (\nabla \cdot \vec{F}) \, dV \] where \(\nabla \cdot \vec{F}\) is the divergence of the vector field.

Step 1: Calculate the Divergence of Vector Field \(\vec{F}\)
The given vector field is: \[ \vec{F} = 4xz\hat{i} - y^2\hat{j} + yz\hat{k} \] The divergence of a vector field \(\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}\) is defined as: \[ \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] Compute each partial derivative: \[ \frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x}(4xz) = 4z \] \[ \frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y}(-y^2) = -2y \] \[ \frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z}(yz) = y \] Summing these derivatives gives: \[ \nabla \cdot \vec{F} = 4z - 2y + y = 4z - y \]

Step 2: Set up and Evaluate the Triple Volume Integral
The volume of the cube is defined by the rectangular boundaries: \( 0 \le x \le 1 \), \( 0 \le y \le 1 \), and \( 0 \le z \le 1 \). Set up the iterated triple integral: \[ \iiint_V (\nabla \cdot \vec{F}) \, dV = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (4z - y) \, dx \, dy \, z \] Since the integrand expression does not explicitly depend on \( x \), integrating with respect to \( x \) over the interval \([0, 1]\) simply yields a factor of 1: \[ \int_{0}^{1} dx = 1 \] The expression simplifies to a double integral: \[ \int_{0}^{1} \int_{0}^{1} (4z - y) \, dy \, dz \] Integrate next with respect to \( y \): \[ \int_{0}^{1} (4z - y) \, dy = \left[ 4zy - \frac{y^2}{2} \right]_{0}^{1} = \left( 4z(1) - \frac{1^2}{2} \right) - 0 = 4z - \frac{1}{2} \] Finally, integrate this result with respect to \( z \): \[ \int_{0}^{1} \left(4z - \frac{1}{2}\right) dz = \left[ 2z^2 - \frac{1}{2}z \right]_{0}^{1} = \left( 2(1)^2 - \frac{1}{2}(1) \right) - 0 = 2 - \frac{1}{2} = \frac{3}{2} \] The volume integral evaluates to \( \frac{3}{2} \), which matches Option (D).