Area swept by $\mathbf{v}(t) - \mathbf{v}(0)$

Here is my numerical attempt. Let the solution at time $t$ be $\vec{r}_1=\vec{v}(t)$ . Let the solution after a time $t + \delta t$ be $\vec{r}_2=\vec{v}(t+\delta t)$ . Using a little bit of imagination, the vectors $\vec{r}_1 -\vec{v}(0)$ , $\vec{r}_2-\vec{v}(0)$ and the elementary arc length vector joining these two vector tips form a triangle. The elementary area of the triangle is the area swept and is given by:

$\delta A_S = \frac{\lvert \left(\vec{r}_1 -\vec{v}(0)\right) \times \left(\vec{r}_2 -\vec{v}(0)\right)\rvert}{2}$ $\delta A_S = \frac{\lvert \left(\vec{v}(t) -\vec{v}(0)\right) \times \left(\vec{v}(t+\delta t)-\vec{v}(0)\right)\rvert}{2}$

From here, I have just written a short script of code for the further evaluation:

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clear all
clc

% The A matrix:
A      = [-0.175 0 0.175;0.35 -0.3 0.25;0.525 0 -0.525];

% Initial conditions:
v(:,1) = [1;2;4];

% Time step and initialisation
dt     = 1e-5;
tf     = 50;
t      = 0:dt:tf;


% Numerical integration of ODE system:
for k = 1:length(t)-1

    v(:,k+1) = v(:,k) + dt*A*v(:,k);

end

% Swept area initialisation:
As = 0;

% Numerical integration to compute swept area:
for k = 1:length(t)-1

    dAs = norm(cross(v(:,k)-v(:,1),v(:,k+1)-v(:,1)))/2;
    As  = As + dAs;
end
ANSWER = As

% ANSWER = 0.3558

As before, the matrix $A$ has eigenvalues $-\tfrac{7}{10}$ , $-\tfrac{3}{10}$ and $0$ , with corresponding eigenvectors $\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) \hspace{1.5cm} \left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) \hspace{1.5cm} \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ Thus $e^{At}\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) = e^{-\frac{7t}{10}}\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) \hspace{1cm} e^{At} \left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) = e^{-\frac{3t}{10}}\left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) \hspace{1cm} e^{At}\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) = \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ Since $\mathbf{v}(0) \; = \; \left(\begin{array}{c}1 \\ 2 \\ 4 \end{array}\right) \; = \; -\tfrac34\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) - \tfrac34 \left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) + \tfrac74\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ we have $\mathbf{v}(t) \; = \; e^{At}\mathbf{v}(0) \; = \; -\tfrac34e^{-\frac{7t}{10}}\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) - \tfrac34 e^{-\frac{3t}{10}}\left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) + \tfrac74\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ The vector $\mathbf{v}(t) - \mathbf{v}(0)$ is always perpendicular to $(3,0,1)^\mathsf{T}$ , and it traces a planar curve, which can be described parametrically in terms of components (calculated by finding two mutually perpendicular unit vectors that are normal to $(3,0,1)^\mathsf{T}$ ): $\begin{aligned} X(t) & = \; \big[\mathbf{v}(t) - \mathbf{v}(0)\big] \cdot \tfrac{1}{\sqrt{14}}\left(\begin{array}{c}1 \\ 2 \\ -3 \end{array} \right) \\[2ex] Y(t) & = \; \big[\mathbf{v}(t) - \mathbf{v}(0)\big] \cdot \tfrac{1}{\sqrt{35}}\left(\begin{array}{c}1 \\ -5 \\ -3 \end{array} \right) \end{aligned}$ Plotting $Y$ parametrically against $X$ gives the shape of the region swept out by $\mathbf{v}(t) - \mathbf{v}(0)$ :

and the area of this region is $\int_0^\infty X'(t)Y(t)\,dt \; = \; \tfrac{9}{80}\sqrt{10} = \boxed{0.355756}$