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

Answer: (A)

$4\pi(a+b)\sqrt{a^2-b^2}$

Answer 2

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