Surface Area — Question 3

Question 3 – Solution

Strategy. The annulus avoids the cone’s nondifferentiable vertex, and the cone has constant slope there.

Step 1: Gradient Write $r=\sqrt{x^2+y^2}$. For $r>0$, $$f_x=\frac{x}{r},\qquad f_y=\frac{y}{r}, \qquad f_x^2+f_y^2=1.$$ The entire projection has $r\ge 1$, so no derivative singularity occurs.

Step 2: Area $$S=\iint_D\sqrt{1+1}\,dA =\sqrt 2\,\pi(3^2-1^2) =\boxed{8\sqrt 2\,\pi}.$$

Verification The frustum has radii $1$ and $3$. Its slant height is $\sqrt{(3-1)^2+(3-1)^2}=2\sqrt 2$, so the classical lateral area $\pi(R+r)\ell=\pi(3+1)(2\sqrt 2)=8\sqrt 2\pi$ agrees.