Question

Consider a set \(S_1 = \{x = (x_1, x_2, x_3)^T \in \mathbb{R}^3 \mid x^Tx \leq 16\}\). Let \(S_2\) be another set which is a subspace of \(\mathbb{R}^3\) with dimension two.
Which of the following gives the area of \(S_1 \cap S_2\)?

• A. \(16\pi\)
• B. \(4\pi\)
• C. \(4\pi^2\)
• D. \(16\pi^2\)

Hint:

\(S_1\) is a solid ball of radius 4 centered at the origin, \(S_2\) is a plane through the origin. Their intersection is a disk of radius 4, so use area \(= \pi r^2\).

Answer 1

The Correct Option is A

Step 1: Identify what \(S_1\) is.
\(x^Tx = x_1^2 + x_2^2 + x_3^2\), which is just the squared distance of the point \(x\) from the origin. So \(S_1 = \{x \in \mathbb{R}^3 \mid x_1^2+x_2^2+x_3^2 \leq 16\}\) is exactly the solid ball of radius \(\sqrt{16} = 4\) centered at the origin.

Step 2: Identify what \(S_2\) is.
A subspace of \(\mathbb{R}^3\) always contains the origin (since it must be closed under scalar multiplication, and multiplying by \(0\) gives the zero vector). A subspace of dimension two is therefore a flat plane passing through the origin.

Step 3: Understand the intersection geometrically.
\(S_1\) is a solid ball centered at the origin with radius \(4\), and \(S_2\) is a plane that passes exactly through that same center (the origin). When a plane cuts through the exact center of a ball, the cross-section formed is a flat disk whose radius equals the ball's own radius, because every point of the plane that is within distance \(4\) of the center (which is also on the plane) satisfies the ball's inequality.

Step 4: State the shape and compute its area.
So \(S_1 \cap S_2\) is a disk of radius \(4\) lying in the plane \(S_2\). The area of a disk of radius \(r\) is \(\pi r^2\):
\[ \text{Area} = \pi (4)^2 = 16\pi \]

Step 5: Why the other options are wrong.
(B) \(4\pi\) would be the area of a disk of radius \(2\), which does not match a ball of radius \(4\).
(C) and (D) involve \(\pi^2\), which does not arise in a plain 2D area formula, these look like errors from squaring \(\pi\) or confusing area with some other quantity like surface area of a sphere in three dimensions.

Final Answer:
\[ \boxed{\text{Area}(S_1 \cap S_2) = 16\pi} \]