To solve the integral of $F(x, y, z) = 4z$ over the portion of the plane $x + y + z = 3$ that lies above the square $0 \leq x \leq 1, 0 \leq y \leq 1$ in the $xy$-plane, we follow these steps:

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

From the plane equation $x + y + z = 3$, solve for $z$: $z = 3 - x - y$

Step 2: Compute the differential element of the plane

The surface is parametrized as: $\vec{r}(x, y) = \langle x, y, 3 - x - y \rangle$ The partial derivatives with respect to $x$ and $y$ are: $\vec{r}_x = \langle 1, 0, -1 \rangle, \quad \vec{r}_y = \langle 0, 1, -1 \rangle$ The cross product $\vec{r}_x \times \vec{r}_y$ gives the normal vector:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{vmatrix} = \mathbf{i} (0 - (-1)) - \mathbf{j} (1 - 0) + \mathbf{k} (1 - 0) = \langle 1, -1, 1 \rangle$$ The magnitude of this vector is: $$|\vec{r}_x \times \vec{r}_y| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$ Thus, the area element is: $$dS = |\vec{r}_x \times \vec{r}_y| \, dx \, dy = \sqrt{3} \, dx \, dy$$ ### Step 3: Write the surface integral The integral is: $$\iint_S F(x, y, z) \, dS = \iint_R F(x, y, 3 - x - y) |\vec{r}_x \times \vec{r}_y| \, dx \, dy$$ Substitute $$F(x, y, z) = 4z$$ and $$z = 3 - x - y$$: $$\iint_S F(x, y, z) \, dS = \iint_R 4(3 - x - y) \sqrt{3} \, dx \, dy$$ Factor out $$\sqrt{3}$$: $$\iint_S F(x, y, z) \, dS = 4\sqrt{3} \iint_R (3 - x - y) \, dx \, dy$$ ### Step 4: Set up the region of integration The region $$R$$ is the square $$0 \leq x \leq 1, 0 \leq y \leq 1$$. ### Step 5: Compute the integral Expand the integral: $$\iint_R (3 - x - y) \, dx \, dy = \int_0^1 \int_0^1 (3 - x - y) \, dx \, dy$$ Evaluate the inner integral with respect to $$x$$: $$\int_0^1 (3 - x - y) \, dx = \left[ 3x - \frac{x^2}{2} - xy \right]_0^1 = (3(1) - \frac{1^2}{2} - (1)y) - (3(0) - \frac{0^2}{2} - (0)y)$$ $$= 3 - \frac{1}{2} - y = \frac{5}{2} - y$$ Now, integrate with respect to $$y$$: $$\int_0^1 \left( \frac{5}{2} - y \right) \, dy = \left[ \frac{5}{2}y - \frac{y^2}{2} \right]_0^1 = \left( \frac{5}{2}(1) - \frac{1^2}{2} \right) - \left( \frac{5}{2}(0) - \frac{0^2}{2} \right)$$ $$= \frac{5}{2} - \frac{1}{2} = 2$$ ### Step 6: Multiply by the constant Multiply the result by $$4\sqrt{3}$$: $$\iint_S F(x, y, z) \, dS = 4\sqrt{3} \cdot 2 = 8\sqrt{3}$$ ### Final Answer: $$\boxed{8\sqrt{3}}$$ Would you like additional explanation on any step? Here are 5 related questions to explore: 1. How would the integral change if the plane equation were different, such as $$x + y + z = 5$$? 2. What is the geometric interpretation of the scalar $$8\sqrt{3}$$? 3. How would you evaluate this integral using cylindrical or spherical coordinates? 4. Can this problem be solved numerically for more complex surfaces? 5. What if the region $$R$$ were not a square but another shape, like a triangle? **Tip**: Always check the bounds of your integration carefully when setting up surface integrals.