Surface Area — Question 9

Question 9 – Solution

Strategy. Although the height changes sign by direction, the squared gradient depends only on distance from the origin.

Step 1: Geometry

Along $y=0$ the trace opens upward, while along $x=0$ it opens downward. Differentiation gives $$f_x=x,\qquad f_y=-y,\qquad \sqrt{1+f_x^2+f_y^2}=\sqrt{1+r^2}.$$

Step 2: Polar integral $$\begin{align*} S&=\int_0^{2\pi}\int_0^{\sqrt 3}\sqrt{1+r^2}\,r\,dr\,d\theta\\ &=2\pi\left[\frac{(1+r^2)^{3/2}}{3}\right]_0^{\sqrt 3}\\ &=\frac{2\pi}{3}(8-1)=\boxed{\frac{14\pi}{3}}. \end{align*}$$

Verification The projected disk has area $3\pi$, while the area factor ranges from $1$ to $2$. Hence $3\pi<S<6\pi$, and $14\pi/3$ lies in that interval.