Evaluate the surface integral of vector A over the plane x+y+z=1 in the first octant. Using spherical coordinates, evaluate the integral of (x²+y²+z²)dxdydz where V is the sphere x²+y²+z²=a² in the first octant.

Triple integral using spherical coordinates

Step 1: The region is $x^2 + y^2 + z^2 \leq 4$, $x \geq 0$, $y \geq 0$, $z \geq 0$. In spherical coordinates, $0 \leq \rho \leq 2$, $0 \leq \theta \leq \pi/2$, $0 \leq \phi \leq \pi/2$. Integrand $(x^2 + y^2 + z^2)^{3/4} = \rho^{3/2}$. $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$.

$$I = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^{3/2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^{7/2} \sin\phi \, d\rho \, d\phi \, d\theta.$$

Step 2: $\int_0^2 \rho^{7/2} \, d\rho = \left[ \frac{2}{9} \rho^{9/2} \right]_0^2 = \frac{2}{9} \cdot 2^{9/2} = \frac{2}{9} \cdot 2^{4 + 1/2} = \frac{2}{9} \cdot 16\sqrt{2} = \frac{32\sqrt{2}}{9}$.

Step 3: $\int_0^{\pi/2} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi/2} = 1$.

Step 4: $\int_0^{\pi/2} d\theta = \pi/2$.

Step 5: $I = \frac{32\sqrt{2}}{9} \cdot 1 \cdot \frac{\pi}{2} = \frac{16\pi\sqrt{2}}{9}$.

\dfrac{16\pi\sqrt{2}}{9}

Eigenvalues of $A = \begin{bmatrix} \frac{1}{2} & 1 \\ -1 & \frac{1}{2} \end{bmatrix}$

Step 1: $A - \lambda I = \begin{bmatrix} \frac{1}{2}-\lambda & 1 \\ -1 & \frac{1}{2}-\lambda \end{bmatrix}$.

$\det(A - \lambda I) = \left(\frac{1}{2} - \lambda\right)^2 - (1)(-1) = \left(\frac{1}{2} - \lambda\right)^2 + 1 = 0$.

Step 2: $\left(\frac{1}{2} - \lambda\right)^2 = -1$, $\frac{1}{2} - \lambda = \pm i$, $\lambda = \frac{1}{2} \mp i$.

\lambda = \dfrac{1}{2} \pm i

Linear system $\frac{dx}{dt} = -5y$, $\frac{dy}{dt} = -5x$