QuestionSuppose that $S$ is a surface in $\mathbf{R}^{3}$ with a parametrization $\Phi$ whose domain $\mathcal{D}$ is the square shown in the figure.
The values of a function $f$, a vector field $\mathbf{F}$, and the normal vector $\mathbf{N}=\mathbf{T}_{u} \times \mathrm{T}_{v}$ at $\Phi(P)$ are given for the four sample points in $D$ in the table. \begin{tabular}{|c|c|c|c|} \hline Point $P$ in $D$ & f & F & N \\ \hline A & 4 & $(2,4,4)$ & (I, I, 1) \\ \hline B & 1 & $(1,1,7)$ & (I, I, 0) \\ \hline C & 2 & $(3,3,-3)$ & $(1,0,-1)$ \\ \hline D & 8 & (0, 1,8) & $(2,1,0)$ \\ \hline \end{tabular}
Estimate the surface integrals of $f$ and $\mathbf{F}$ over $S$. (Use decimal notation. Round your answers to three decimal places.) $$\iint_{S} f(x, y, z) d S \approx$$ $\square$ $$\iint_{S} \mathbf{F} \cdot d \mathbf{S} \approx$$ $\square$

Studdy Solution
Estimate the surface integral of $\mathbf{F} \cdot \mathbf{N}$ over $S$ using the midpoint rule. Calculate the average value of $\mathbf{F} \cdot \mathbf{N}$ at the sample points and multiply by the area of the domain $\mathcal{D}$.
- Dot products $\mathbf{F} \cdot \mathbf{N}$ at points: - $A: (2,4,4) \cdot (1,1,1) = 2 + 4 + 4 = 10$ - $B: (1,1,7) \cdot (1,1,0) = 1 + 1 + 0 = 2$ - $C: (3,3,-3) \cdot (1,0,-1) = 3 + 0 + 3 = 0$ - $D: (0,1,8) \cdot (2,1,0) = 0 + 1 + 0 = 1$ - Average value of $\mathbf{F} \cdot \mathbf{N}$: \frac{10 + 2 + 0 + 1}{4} = \frac{13}{4} = 3.25 \] - Estimate of \(\iint_{S} \mathbf{F} \cdot d\mathbf{S}\): 3.25 \times 1 = 3.25 \]
The estimated surface integrals are: $\iint_{S} f(x, y, z) dS \approx 3.750$ $\iint_{S} \mathbf{F} \cdot d\mathbf{S} \approx 3.250$