Question

The value of \( \oint_S \vec{F} \cdot d\vec{s} \) where \( \vec{F} = 4x\hat{i} - 2y^2\hat{j} + z^2\hat{k} \) taken over the cylinder \( x^2+y^2=4, z=0 \) and \( z=3 \) is:

• A. 126\(\pi\)
• B. 168\(\pi\)
• C. 42\(\pi\)
• D. 84\(\pi\)

Hint:

When asked to evaluate a surface integral over a simple closed surface (like a sphere, cylinder, or cube), always check if the Divergence Theorem can be applied first. It often simplifies the problem from multiple surface integrals to a single, often easier, volume integral. Also, look for symmetries that might make parts of the integral zero, like integrating \(y\) or \(sin\theta\) over a symmetric domain.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem asks for the surface integral of a vector field \( \vec{F} \) over a closed surface S, which is the total flux of \( \vec{F} \) out of the surface. The surface is a closed cylinder. The Divergence Theorem (also known as Gauss's theorem) is the ideal tool for this, as it relates the surface integral (flux) to a volume integral of the divergence of the field.
Step 2: Key Formula or Approach:
The Divergence Theorem states: \[ \oint_S \vec{F} \cdot d\vec{s} = \iiint_V (\nabla \cdot \vec{F}) \, dV \] where V is the volume enclosed by the closed surface S. First, we need to calculate the divergence of \( \vec{F} \). \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(4x) + \frac{\partial}{\partial y}(-2y^2) + \frac{\partial}{\partial z}(z^2) \] Step 3: Detailed Explanation:
1. Calculate the Divergence:
\[ \nabla \cdot \vec{F} = 4 - 4y + 2z \] 2. Set up the Volume Integral:
The volume V is a cylinder with radius \(r=2\) and height from \(z=0\) to \(z=3\). The integral is: \[ \iiint_V (4 - 4y + 2z) \, dV \] It's best to evaluate this in cylindrical coordinates. In cylindrical coordinates: \( x = r\cos\theta, y = r\sin\theta, z = z \), and \( dV = r \, dz \, dr \, d\theta \). The limits are: \( 0 \le \theta \le 2\pi \), \( 0 \le r \le 2 \), \( 0 \le z \le 3 \). The integral becomes: \[ \int_0^{2\pi} \int_0^2 \int_0^3 (4 - 4r\sin\theta + 2z) \, r \, dz \, dr \, d\theta \] 3. Evaluate the Integral:
We can separate the integral into three parts: \[ \int_0^{2\pi} \int_0^2 \int_0^3 4r \, dz \, dr \, d\theta - \int_0^{2\pi} \int_0^2 \int_0^3 4r^2\sin\theta \, dz \, dr \, d\theta + \int_0^{2\pi} \int_0^2 \int_0^3 2zr \, dz \, dr \, d\theta \] Part 1: \( \int_0^{2\pi} d\theta \int_0^2 r dr \int_0^3 4 dz = (2\pi) \times [\frac{r^2}{2}]_0^2 \times [4z]_0^3 = 2\pi \times 2 \times 12 = 48\pi \). Part 2: The integral with respect to \(\theta\) is \( \int_0^{2\pi} \sin\theta \, d\theta = [-\cos\theta]_0^{2\pi} = -1 - (-1) = 0 \). So this entire part is 0. Part 3: \( \int_0^{2\pi} d\theta \int_0^2 r dr \int_0^3 2z dz = (2\pi) \times [\frac{r^2}{2}]_0^2 \times [z^2]_0^3 = 2\pi \times 2 \times 9 = 36\pi \). Total Value: The total value of the integral is the sum of the parts: \[ 48\pi - 0 + 36\pi = 84\pi \] Step 4: Final Answer:
The value of the surface integral is \(84\pi\).