Compute the flux of the vector field F⃗ (x,y,z)=6i⃗ +4j⃗ +5k⃗
through the rectangular region with corners at (1,1,0)
, (0,1,0)
, (0,0,2)
, and (1,0,2)
oriented in the positive z-direction

To compute the flux of the vector field \(\mathbf{F}(x,y,z) = 6\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\) through the given rectangular region, we follow these steps:

### Step-by-Step Solution

1. **Identify the surface and orientation:**
   The rectangular region has corners at \((1,1,0)\), \((0,1,0)\), \((0,0,2)\), and \((1,0,2)\). This region lies on a plane parallel to the x-y plane, but it's skewed in the z-direction.

2. **Parameterize the surface:**
   The surface can be parameterized using two parameters \(u\) and \(v\). Let:
   \[
   \mathbf{r}(u,v) = (u, v, 2u)
   \]
   where \(0 \leq u \leq 1\) and \(0 \leq v \leq 1\).

3. **Compute the tangent vectors:**
   \[
   \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = (1, 0, 2)
   \]
   \[
   \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = (0, 1, 0)
   \]

4. **Compute the normal vector:**
   The normal vector \(\mathbf{n}\) is given by the cross product of \(\mathbf{r}_u\) and \(\mathbf{r}_v\):
   \[
   \mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   1 & 0 & 2 \\
   0 & 1 & 0 
   \end{vmatrix} = (0 \cdot 0 - 2 \cdot 1)\mathbf{i} - (1 \cdot 0 - 0 \cdot 0)\mathbf{j} + (1 \cdot 1 - 0 \cdot 0)\mathbf{k} = -2\mathbf{i} + 0\mathbf{j} + 1\mathbf{k}
   \]
   Thus, \(\mathbf{n} = -2\mathbf{i} + \mathbf{k}\).

5. **Normalize the normal vector:**
   The normal vector \(\mathbf{n}\) is already perpendicular to the surface. To ensure it points in the positive z-direction, we use \(\mathbf{n} = -2\mathbf{i} + \mathbf{k}\).

6. **Compute the flux:**
   The flux \(\Phi\) through the surface is given by:
   \[
   \Phi = \iint_S \mathbf{F} \cdot d\mathbf{S}
   \]
   where \(d\mathbf{S} = \mathbf{n} \, du \, dv\).
   
   Substituting \(\mathbf{F}\) and \(\mathbf{n}\):
   \[
   \mathbf{F} \cdot \mathbf{n} = (6\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) \cdot (-2\mathbf{i} + \mathbf{k}) = 6(-2) + 4(0) + 5(1) = -12 + 5 = -7
   \]

7. **Set up the integral:**
   \[
   \Phi = \iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^1 \int_0^1 -7 \, du \, dv
   \]

8. **Evaluate the integral:**
   \[
   \Phi = \int_0^1 \int_0^1 -7 \, du \, dv = -7 \int_0^1 \int_0^1 du \, dv = -7 \left( \int_0^1 du \right) \left( \int_0^1 dv \right) = -7 (1)(1) = -7
   \]

### Final Answer

The flux of the vector field \(\mathbf{F}(x,y,z) = 6\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\) through the given rectangular region is \(-7\).

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Do you want more details or have any questions about this problem?

Here are some related questions you might find interesting:

1. How to compute the flux of a vector field through a more complex surface?
2. What are the applications of flux in physics and engineering?
3. How do you parameterize a surface in different coordinate systems?
4. What is the divergence theorem and how is it applied?
5. How does the orientation of a surface affect the flux calculation?
6. What are some common mistakes to avoid when computing surface integrals?
7. How do you find the normal vector to a surface given its parameterization?
8. Can you explain the physical meaning of a negative flux value?

**Tip:** Always check the orientation of your surface normal when computing flux; it ensures the correct sign and direction for your integral result.

Learn how to compute the flux of the vector field F(x,y,z)=6i + 4j + 5k through a rectangular region with corners at (1,1,0), (0,1,0), (0,0,2), and (1,0,2) oriented in the positive z-direction. Follow a step-by-step solution including surface parameterization, computation of normal vector, and surface integral. Suitable for college-level mathematics students.

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