Surface Area — Question 5

Question 5 – Solution

Strategy. Project onto the $yz$-plane, where both the region and squared gradient are radial.

Step 1: Projection formula For $x=g(y,z)$, $$dS=\sqrt{1+g_y^2+g_z^2}\,dy\,dz.$$ Here $g=y^2+z^2$, so with $s^2=y^2+z^2$ the factor is $\sqrt{1+4s^2}$.

Step 2: Evaluate $$\begin{align*} S&=\int_0^{2\pi}\int_0^1\sqrt{1+4s^2}\,s\,ds\,d\theta\\ &=2\pi\left[\frac{(1+4s^2)^{3/2}}{12}\right]_0^1 =\boxed{\frac{\pi}{6}(5\sqrt 5-1)}. \end{align*}$$

Verification The graph projects one-to-one onto a disk of area $\pi$, and the area factor exceeds $1$ except at its center. Accordingly, the computed surface area is strictly larger than $\pi$.