Surface Area — Question 4

Question 4 – Solution

Strategy. Integrate over disks of radius $b<3$, then take the limit as $b\to 3^-$.

Step 1: Area factor

With $r^2=x^2+y^2$, $$f_x=-\frac{x}{\sqrt{9-r^2}},\qquad f_y=-\frac{y}{\sqrt{9-r^2}},$$ so $$\sqrt{1+f_x^2+f_y^2}=\frac{3}{\sqrt{9-r^2}}.$$

Step 2: Improper integral $$\begin{align*} S&=\lim_{b\to 3}\int_0^{2\pi}\int_0^b \frac{3r}{\sqrt{9-r^2}}\,dr\,d\theta\\ &=\lim_{b\to 3}6\pi\left(3-\sqrt{9-b^2}\right) =\boxed{18\pi}. \end{align*}$$

Verification The factor diverges as $r\to 3^-$ because the sphere becomes vertical, but its antiderivative has a finite limit. The result also equals half the full sphere area $4\pi(3^2)=36\pi$.