Identify the Geometric Shapes:

• The set $S_1 = \{x = (x_1, x_2, x_3)^T \in \mathbb{R}^3 \mid x^T x \leq 16\}$ represents a solid ball in 3D space centered at the origin. The inequality $x^T x \leq 16$ implies $||x||^2 \leq 16$, meaning the radius $R$ satisfies $R^2 = 16$, so $R=4$.
• The set $S_2$ is a subspace of $\mathbb{R}^3$ with dimension two. This corresponds to a plane passing through the origin in $\mathbb{R}^3$.

Determine the Intersection:

• The intersection $S_1 \cap S_2$ is the set of points that lie both within the solid ball $S_1$ and on the plane $S_2$.
• Geometrically, this intersection forms a disk (a flat circle including its interior) lying within the plane $S_2$.
• Since the plane $S_2$ passes through the center of the ball $S_1$, the radius of this disk is the same as the radius of the ball, which is $R=4$.

Calculate the Area:

• The area of a disk with radius $r$ is given by the formula: $A = \pi r^2$
• In this case, the radius $r = 4$. Substituting this value into the formula: $A = \pi (4)^2$ $A = 16\pi$

The area of the intersection $S_1 \cap S_2$ is $16\pi$. This corresponds to Option A.