Surface Integrals — Question 4

Question 4 – Solution

Strategy. Use height as the radius parameter. The factor of $u$ in $dS$ cancels against the weight $1/z$ in the second integral.

Step 1: Parametrize $$\mathbf r(u,v)=\langle u\cos v,u\sin v,u\rangle, \quad 1\le u\le 3,\quad 0\le v\le 2\pi.$$ The tangent vectors have cross-product magnitude $$\lVert\mathbf r_u\times\mathbf r_v\rVert=u\sqrt 2,$$ so $\boxed{dS=u\sqrt 2\,du\,dv}$.

Step 2: Area $$\operatorname{area}(S)=\int_0^{2\pi}\int_1^3u\sqrt 2\,du\,dv =\boxed{8\pi\sqrt 2}.$$

Step 3: Weighted integral Since $z=u$, $$\iint_S\frac 1z\,dS =\int_0^{2\pi}\int_1^3\frac 1u u\sqrt 2\,du\,dv =\boxed{4\pi\sqrt 2}.$$ Its average value is therefore $$\boxed{\frac{4\pi\sqrt 2}{8\pi\sqrt 2}=\frac 12}.$$

Verification On the frustum, $1/3\le 1/z\le 1$; the average $1/2$ lies in this interval.