Question

If \[ \mathbf{F}=ax\mathbf{i}+by\mathbf{j}+cz\mathbf{k}, \] where \(a\), \(b\), and \(c\) are constants, and \(S\) is the surface of the unit sphere, then \[ \iint_{S}\mathbf{F}\cdot\mathbf{n}\,dS=\_ \]

• A. $4\pi/3(a + b + c)$
• B. $4\pi(a + b + c)$
• C. $2\pi/3(a + b + c)$
• D. $\pi(a + b + c)$

Hint:

Use Gauss Divergence Theorem to convert a complex surface integral into a simple volume integral whenever the surface is closed.

Answer 1

The Correct Option is A

Step 1: Concept
By Gauss Divergence Theorem, $\iint_{S} F \cdot n dS = \iiint_{V} (\nabla \cdot F) dV$.

Step 2: Meaning
Calculate $\nabla \cdot F = \frac{\partial(ax)}{\partial x} + \frac{\partial(by)}{\partial y} + \frac{\partial(cz)}{\partial z} = a + b + c$.

Step 3: Analysis
Since $(a + b + c)$ is constant, the integral becomes $(a + b + c) \iiint_{V} dV$, where $V$ is the volume of the unit sphere.

Step 4: Conclusion
The volume of a unit sphere ($r=1$) is $\frac{4}{3}\pi(1)^{3} = \frac{4}{3}\pi$. Thus, the integral is $\frac{4}{3}\pi(a + b + c)$. Final Answer: (A)