Evaluate A · n d S, where A=(x+y) i+(y+z) j+x y z k and S is the surface of the plane x+y+z=1 in the first octant.

Surface element: $dS = \rho \sin\phi \, d\rho \, d\theta$.

$\int_0^{\pi/2} \sin\theta \, d\theta = \left[ -\cos\theta \right]_0^{\pi/2} = 0 - (-1) = 1$,

so $\int_0^{\pi/2} (\cos\theta - \sin\theta) \, d\theta = 1 - 1 = 0$.

Region $V$: $0 \leq \theta \leq \pi/2$, $0 \leq \phi \leq \pi/4$, $0 \leq \rho \leq 1$.

$\int_0^1 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^1 = 1/4$.

Characteristic equation: $\det(A - \lambda I) = (1 - \lambda)^2 = 0$, so $\lambda = 1$ (multiplicity 2).

$(A - I)\mathbf{v} = \mathbf{0}$ gives zero matrix equation, eigenspace is all $\mathbb{R}^2$.

Basis: $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

\lambda=1 (mult. 2),\ eigenspace\ \mathbb{R}^2

$\det(V - \lambda I) = (-1 - \lambda)(1 - \lambda) = 0$, so $\lambda_1 = -1$, $\lambda_2 = 1$.

$(V + I)\mathbf{v} = \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix} \mathbf{v} = \mathbf{0}$, so $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

For $\lambda_2 = 1$: $(V - I)\mathbf{v} = \begin{bmatrix} -2 & 0 \\ 0 & 0 \end{bmatrix} \mathbf{v} = \mathbf{0}$, so $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

\lambda_1=-1\ ( \begin{bmatrix} 1 \ 0 \end{bmatrix} ),\ \lambda_2=1\ ( \begin{bmatrix} 0 \ 1 \end{bmatrix} )

\Phi(t)=\begin{bmatrix} e^{-t} & 0 \ 0 & e^{t} \end{bmatrix}

$x = r \cos\theta$, $y = r \sin\theta$, $dA = r \, dr \, d\theta$, $\theta: 0 \to \pi$, $r: 0 \to 1$.

$\int_0^\pi \cos\theta \, d\theta = [\sin\theta]_0^\pi = 0$,

$\int_0^\pi \sin\theta \, d\theta = [-\cos\theta]_0^\pi = 1 - (-1) = 2$,

so $2 \cdot (0 + 2) \cdot \frac{1}{3} = 2 \cdot 2 \cdot \frac{1}{3} = \frac{4}{3}$.

\dfrac{4}{3}