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 correct option is (A) : \(4\pi(a+b)\sqrt{a^2-b^2}\).