To find the surface integral of the vector field $\mathbf{F}(x,y,z) = -\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$ across the rectangular surface $z=0$, with $0 \leq x \leq 3$ and $0 \leq y \leq 3$, in the $\mathbf{k}$ direction, we follow these steps:

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1. Surface Integral Definition:

The surface integral of a vector field $\mathbf{F}$ across a surface $S$ is given by:

$\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$

Where:

• $\mathbf{n}$ is the unit normal vector to the surface.
• $dS$ is the infinitesimal surface area element.

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2. Surface Properties:

• The surface lies in the plane $z=0$, so the normal vector $\mathbf{n}$ is in the $\mathbf{k}$ direction.
• The surface is parameterized by $x$ and $y$: $(x, y, z=0)$.
• The unit normal vector is $\mathbf{n} = \mathbf{k}$.
• The surface area element $dS$ in this case is simply $dx \, dy$ because the surface lies in the $xy$-plane.

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3. Dot Product:

The vector field is $\mathbf{F} = -\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$.

Taking the dot product with the unit normal vector $\mathbf{n} = \mathbf{k}$:

$\mathbf{F} \cdot \mathbf{n} = (-\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) \cdot \mathbf{k} = 3$

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4. Evaluate the Integral:

The surface integral simplifies to:

$\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iint_R 3 \, dx \, dy$

Where $R$ is the region $0 \leq x \leq 3$, $0 \leq y \leq 3$.

This becomes:

$\iint_R 3 \, dx \, dy = 3 \int_0^3 \int_0^3 dx \, dy$

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5. Compute the Double Integral:

$\int_0^3 \int_0^3 dx \, dy = \int_0^3 \left[ \int_0^3 dx \right] dy = \int_0^3 (3) \, dy = 3 \cdot 3 = 9$

Thus, the surface integral is:

$\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = 3 \cdot 9 = 27$

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Final Answer:

$\boxed{27}$

Let me know if you would like a more detailed explanation of any step!

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Related Questions:

1. How do surface integrals change when the normal vector direction is reversed?
2. What is the physical significance of a surface integral in vector fields?
3. How does parameterization affect the computation of surface integrals?
4. Can this integral be extended to curved surfaces, and how?
5. How would the solution change if the surface was in the plane $z=5$?

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Tip:

When computing surface integrals, always ensure the normal vector is properly aligned with the orientation of the surface to avoid sign errors.