Question

Evaluate \(\iint_S \vec{A}\cdot \hat{n}\,dS\), where \(\vec{A}=2y\hat{i}+yz\hat{j}+xz\hat{k}\) and \(S\) is the surface of the region bounded by \(x=0,\ y=0,\ z=0,\ y=3\) and \(x+2z=6\).

• A. \(45\)
• B. \(81\)
• C. \(135\)
• D. \(351\)

Hint:

For closed surface flux, use divergence theorem: surface integral becomes volume integral of divergence.

Answer 1

The Correct Option is B

Concept:
Use Gauss divergence theorem: \[ \iint_S \vec{A}\cdot \hat{n}\,dS=\iiint_V \vec{\nabla}\cdot \vec{A}\,dV \]

Step 1: Write the vector field. \[ \vec{A}=2y\hat{i}+yz\hat{j}+xz\hat{k} \]

Step 2: Find divergence. \[ \vec{\nabla}\cdot\vec{A} = \frac{\partial}{\partial x}(2y) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(xz) \] \[ \vec{\nabla}\cdot\vec{A} = 0+z+x \] \[ \vec{\nabla}\cdot\vec{A}=x+z \]

Step 3: Define the region.
The region is bounded by: \[ x=0,\quad y=0,\quad z=0,\quad y=3,\quad x+2z=6 \] So: \[ 0\le y\le 3 \] For the \(xz\)-plane: \[ 0\le z\le 3 \] \[ 0\le x\le 6-2z \]

Step 4: Set up the integral. \[ \iiint_V (x+z)\,dV = \int_0^3\int_0^3\int_0^{6-2z}(x+z)\,dx\,dz\,dy \]

Step 5: Integrate with respect to \(x\). \[ \int_0^{6-2z}(x+z)\,dx = \left[\frac{x^2}{2}+zx\right]_0^{6-2z} \] \[ = \frac{(6-2z)^2}{2}+z(6-2z) \]

Step 6: Complete the integration. \[ \int_0^3\int_0^3\left[\frac{(6-2z)^2}{2}+z(6-2z)\right]dz\,dy \] After simplification and integration: \[ =81 \] \[ \therefore \text{Correct Answer is (B)} \]