Surface Integrals — Question 6

Question 6 – Solution

Strategy. The graph factor becomes unbounded at the equator, but multiplication by $z$ cancels it exactly in the interior.

Step 1: Graph factor Let $g(x,y)=\sqrt{9-x^2-y^2}$ over the disk $D:x^2+y^2\le 9$. Since $$g_x=-\frac{x}{z},\qquad g_y=-\frac{y}{z},$$ we have, for $z>0$, $$\begin{align*} \sqrt{1+g_x^2+g_y^2} &=\sqrt{1+\frac{x^2+y^2}{z^2}} =\frac{3}{z}. \end{align*}$$ Thus $dS=(3/z)dA$ away from the boundary.

Step 2: Cancel and integrate The weighted element is simply $$z\,dS=z\frac 3z\,dA=3\,dA.$$ Therefore $$\boxed{\iint_S z\,dS=3\operatorname{area}(D)=27\pi}.$$ The hemisphere has area $2\pi(3^2)=18\pi$, so $$\boxed{\operatorname{Avg}_S(z)=\frac{27\pi}{18\pi}=\frac 32}.$$

Verification Integrating over disks of radius $R<3$ gives $3\pi R^2$; taking $R\to 3^-$ justifies the boundary limit and yields $27\pi$.