Surface Integrals — Question 8

Question 8 – Solution

Strategy. Compute the cross product in the given parameters; its magnitude depends only on $u$, so the $v$-integration separates.

Step 1: Surface element $$\mathbf r_u=\langle\cos v,\sin v,0\rangle, \qquad \mathbf r_v=\langle-u\sin v,u\cos v,1\rangle.$$ Thus $$\mathbf r_u\times\mathbf r_v=\langle\sin v,-\cos v,u\rangle, \qquad \boxed{dS=\sqrt{1+u^2}\,du\,dv}.$$

Step 2: Integrate $$\begin{align*} \operatorname{area}(S) &=2\pi\int_0^1\sqrt{1+u^2}\,du\\ &=2\pi\left[\frac 12\left(u\sqrt{1+u^2} +\ln\bigl(u+\sqrt{1+u^2}\bigr)\right)\right]_0^1\\ &=\boxed{\pi\left(\sqrt 2+\ln(1+\sqrt 2)\right)}. \end{align*}$$

Step 3: Check The cross product never vanishes because its first two components have squared sum $1$. Also $1\le\sqrt{1+u^2}\le\sqrt 2$, so $$2\pi\le\operatorname{area}(S)\le 2\pi\sqrt 2,$$ and the exact result lies between these bounds.