Question

The value of the surface integral \( \iint_S (x^2 \,dydz + y^2 \,dzdx + z^2 \,dxdy) \) over the surface of the cube given by \( 0 \le x \le 2, 0 \le y \le 2, 0 \le z \le 2 \), is

• A. 12
• B. 24
• C. 36
• D. 48

Hint:

The Divergence Theorem is almost always the easiest way to evaluate a flux integral over a simple closed surface like a sphere or a cube. If the problem asks for \( \iint_S \vec{F} . d\vec{S} \) over a closed surface, your first thought should be to calculate the divergence of \( \vec{F} \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem asks to evaluate a surface integral of a vector field over a closed surface (a cube). This is a classic application for the Gauss Divergence Theorem, which converts a surface integral into a simpler volume integral.
Step 2: Key Formula or Approach:
The given surface integral is in the form \( \iint_S \vec{F} . d\vec{S} \), where the vector field is \( \vec{F} = P\hat{i} + Q\hat{j} + R\hat{k} \) and \( d\vec{S} = dydz\hat{i} + dzdx\hat{j} + dxdy\hat{k} \). From the given integral, we can identify the components of the vector field \( \vec{F} \):
- \( P = x^2 \)
- \( Q = y^2 \)
- \( R = z^2 \)
So, \( \vec{F} = x^2\hat{i} + y^2\hat{j} + z^2\hat{k} \).
The Gauss Divergence Theorem states: \[ \iint_S \vec{F} . d\vec{S} = \iiint_V (\nabla . \vec{F}) \,dV \] where V is the volume enclosed by the surface S. The divergence is \( \nabla . \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).
Step 3: Detailed Calculation:
1. Calculate the divergence of \( \vec{F} \): \[ \nabla . \vec{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z = 2(x+y+z) \] 2. Set up the volume integral: The volume V is the cube defined by \( 0 \le x \le 2, 0 \le y \le 2, 0 \le z \le 2 \). The integral becomes: \[ \iiint_V 2(x+y+z) \,dV = 2 \int_{0}^{2} \int_{0}^{2} \int_{0}^{2} (x+y+z) \,dx \,dy \,dz \] 3. Evaluate the integral: Due to symmetry, we can calculate the integral for one variable and multiply by 3. \[ \int_{0}^{2} \int_{0}^{2} \int_{0}^{2} x \,dx \,dy \,dz = \left( \int_{0}^{2} x \,dx \right) \left( \int_{0}^{2} dy \right) \left( \int_{0}^{2} dz \right) \] \[ = \left[ \frac{x^2}{2} \right]_{0}^{2} \times [y]_{0}^{2} \times [z]_{0}^{2} = \left(\frac{4}{2}\right) \times (2) \times (2) = 2 \times 2 \times 2 = 8 \] So, the integral of \(x\) over the volume is 8. By symmetry, the integral of \(y\) and \(z\) over the volume will also be 8. \[ \iiint_V (x+y+z) \,dV = \iiint_V x \,dV + \iiint_V y \,dV + \iiint_V z \,dV = 8 + 8 + 8 = 24 \] Finally, multiply by the factor of 2 from the divergence: \[ \text{Value of integral} = 2 \times 24 = 48 \] Step 4: Final Answer:
The value of the surface integral is 48.