Question

If \( \vec{F} = x^2 \hat{i} + z \hat{j} + yz \hat{k} \), for \( (x, y, z) \in \mathbb{R}^3 \), then:

Evaluate \( \oiint_S \vec{F} \cdot d\vec{S} \), where \( S \) is the surface of the cube formed by \( x = \pm 1, y = \pm 1, z = \pm 1 \):

• A. 24
• B. 6
• C. 1
• D. 0

Hint:

When integrating an odd function (like \(f(x)=x\)) over a symmetric interval (like \([-a, a]\)), the result is always zero. This is a very useful shortcut for evaluating integrals over symmetric domains like cubes or spheres centered at the origin.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem asks for the flux of a vector field through a closed surface (a cube). This is a classic application of the Gauss Divergence Theorem, which relates a surface integral over a closed surface to a volume integral over the region enclosed by that surface.

Step 2: Key Formula or Approach:
The Gauss Divergence Theorem states: \[ \oiint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F}) dV \] Where \( V \) is the volume enclosed by the surface \( S \). The given vector field is \( \vec{F} = x^2 \hat{i} + z \hat{j} + yz \hat{k} \).

Step 3: Detailed Explanation:
First, we calculate the divergence of \( \vec{F} \): \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(z) + \frac{\partial}{\partial z}(yz) \] \[ \nabla \cdot \vec{F} = 2x + 0 + y = 2x + y \] Now, we apply the Divergence Theorem. The volume \( V \) is the cube defined by \( -1 \le x \le 1 \), \( -1 \le y \le 1 \), and \( -1 \le z \le 1 \). \[ \oiint_S \vec{F} \cdot d\vec{S} = \iiint_V (2x + y) dV = \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (2x + y) dz \, dy \, dx \] We can separate the integral into two parts: \[ \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} 2x \, dz \, dy \, dx + \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} y \, dz \, dy \, dx \] For the first part, integrating with respect to \( x \): \( \int_{-1}^{1} 2x \, dx = [x^2]_{-1}^{1} = (1)^2 - (-1)^2 = 1 - 1 = 0 \). Since one of the iterated integrals is 0, the entire first term is 0. For the second part, integrating with respect to \( y \): \( \int_{-1}^{1} y \, dy = [\frac{y^2}{2}]_{-1}^{1} = \frac{1^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0 \). Since one of the iterated integrals is 0, the entire second term is 0. Therefore, the total integral is \( 0 + 0 = 0 \).

Step 4: Final Answer:
The value of the surface integral is 0.