Surface Area — Question 7

Question 7 – Solution

Strategy. Differentiate first: the two gradient components exchange $x$ and $y$, so their squares combine to $r^2$.

Step 1: Gradient and projection $$f_x=y,\qquad f_y=x,\qquad \sqrt{1+f_x^2+f_y^2}=\sqrt{1+x^2+y^2}=\sqrt{1+r^2}.$$

The quarter-disk has $0\le\theta\le\pi/2$ and $0\le r\le 1$.

Step 2: Evaluate $$\begin{align*} S&=\int_0^{\pi/2}\int_0^1\sqrt{1+r^2}\,r\,dr\,d\theta\\ &=\frac{\pi}{2}\left[\frac{(1+r^2)^{3/2}}{3}\right]_0^1 =\boxed{\frac{\pi}{6}(2\sqrt 2-1)}. \end{align*}$$

Verification The angular width is one quarter of a full turn. The factor ranges from $1$ to $\sqrt 2$, so the result must lie between the projection area $\pi/4$ and $\sqrt 2\,\pi/4$; it does.