Question

For a > b > 0, consider
\(D=\left\{(x,y,z) \in \R^3 :x^2+y^2+z^2 \le a^2\ \text{and } x^2+y^2 \ge b^2\right\}.\)
Then, the surface area of the boundary of the solid D is

• A. \(4\pi(a+b)\sqrt{a^2-b^2}\)
• B. \(4\pi(a^2-b\sqrt{a^2-b^2})\)
• C. \(4\pi(a-b)\sqrt{a^2-b^2}\)
• D. \(4\pi(a^2+b\sqrt{a^2-b^2})\)

Answer 1

The Correct Option is A

The region \( D \) is described as the part of a sphere of radius \( a \) and centered at the origin, with the additional restriction that \( x^2 + y^2 \geq b^2 \), which means that the solid is bounded by a cylindrical hole along the \( z \)-axis. To find the surface area of the boundary of the solid \( D \), we need to calculate the surface area of the sphere excluding the portion inside the cylinder. The surface area of the sphere is given by \( 4\pi a^2 \). The region inside the cylinder can be described by the equation \( x^2 + y^2 = b^2 \), so the area of the spherical cap inside the cylinder is given by the formula for the surface area of a spherical zone, \( 2\pi (a^2 - b^2) \). Thus, the surface area of the boundary of the solid \( D \) is the difference between the surface area of the entire sphere and the area inside the cylindrical hole, which is: \[ 4\pi (a + b)\sqrt{a^2 - b^2}. \] Therefore, the correct answer is (A): \( 4\pi(a + b)\sqrt{a^2 - b^2} \).