To solve the given problem, we integrate the function $F(x, y, z) = 2z$ over the portion of the plane $x + y + z = 6$ above the square $0 \leq x \leq 1$ and $0 \leq y \leq 1$ in the $xy$-plane.

Step 1: Express $z$ in terms of $x$ and $y$

From the plane equation $x + y + z = 6$, solve for $z$: $z = 6 - x - y$

Step 2: Set up the integral

The given square in the $xy$-plane determines the bounds for $x$ and $y$. For the square: $0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq 1$

The integrand is $2z$, and $z = 6 - x - y$. Substituting $z$, the integrand becomes: $2z = 2(6 - x - y) = 12 - 2x - 2y$

The integral over the specified region is: $\int_0^1 \int_0^1 (12 - 2x - 2y) \, dy \, dx$

Step 3: Evaluate the integral

Inner integral (with respect to $y$):

$\int_0^1 (12 - 2x - 2y) \, dy = \left[ 12y - 2xy - y^2 \right]_0^1$ Evaluate at the bounds $y = 1$ and $y = 0$: [ \left[ 12(1) - 2x(1) - (1)^2 \right] - \left[ 12(0) - 2x(0) - (0)^2 \right] = 12 - 2x - 1 = 11 - 2x ]

Outer integral (with respect to $x$):

$\int_0^1 (11 - 2x) \, dx = \left[ 11x - x^2 \right]_0^1$ Evaluate at the bounds $x = 1$ and $x = 0$: [ \left[ 11(1) - (1)^2 \right] - \left[ 11(0) - (0)^2 \right] = 11 - 1 = 10 ]

Final Answer:

$\int \int 2z \, dA = 10$

Would you like further clarification or steps elaborated?

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Related Questions:

1. How would the integral change if the plane were $x + y + z = 10$?
2. What happens if the square is expanded to $0 \leq x \leq 2$ and $0 \leq y \leq 2$?
3. How do you parameterize a surface integral in general?
4. What are the physical interpretations of surface integrals in vector fields?
5. How does the orientation of the surface affect the calculation of surface integrals?

Tip:

When solving surface integrals, always ensure the limits of integration match the projection of the surface onto the chosen plane.