Surface Area — Question 2

Question 2 – Solution

Strategy. The squared gradient depends only on $r$, so polar coordinates reduce the area integral to one variable.

Step 1: Geometry and gradient

For $f=x^2+y^2$, $$f_x=2x,\qquad f_y=2y,\qquad \sqrt{1+f_x^2+f_y^2}=\sqrt{1+4r^2}.$$

Step 2: Evaluate $$\begin{align*} S&=\int_0^{2\pi}\int_0^2\sqrt{1+4r^2}\,r\,dr\,d\theta\\ &=2\pi\left[\frac{(1+4r^2)^{3/2}}{12}\right]_0^2 =\boxed{\frac{\pi}{6}\left(17\sqrt{17}-1\right)}. \end{align*}$$

Verification The area factor is at least $1$, so $S$ must exceed the projected disk area $4\pi$. The exact result does. Its integrand also increases with $r$, matching the increasing steepness of the paraboloid.