Problem 1
1. Calculate \(\displaystyle\int\!\!\int_S \sqrt{x^2+y^2+1}\, dS\) where \(S\) is the helicoid given parametrically by \(G(r,\theta) = (r\cos\theta, r\sin\theta, \theta)\), with \(0\le r\le 1\) and \(0\le \theta\le 2\pi\). What is the surface area of \(S\)? (5 points)
Solution. As this is a surface integral, we begin with finding the magnitude of the normal to the surface. Note \(\mathbf{T}_r = (\cos\theta, \sin\theta, 0)\), \(\mathbf{T}_\theta = (-r\sin\theta, r\cos\theta, 1)\), so
\[ \mathbf{T}_r\times\mathbf{T}_\theta = \begin{vmatrix} \vi & \vj & \vk \\ \cos\theta & \sin\theta & 0 \\ -r\sin\theta & r\cos\theta & 1 \end{vmatrix} = (\sin\theta, -\cos\theta, r). \]
Lastly, note that \(||\mathbf{T}_r\times\mathbf{T}_\theta|| = \sqrt{1+r^2}\). We proceed to the integral calculation itself. One has \[ \begin{align*} \int\!\!\int_S \sqrt{x^2+y^2+1}\, dS &= \int_0^{2\pi}\!\int_0^1 \left(\sqrt{r^2\cos^2\theta+r^2\sin^2\theta+1}\right)||\mathbf{T}_r\times\mathbf{T}_\theta||\, dr\, d\theta = \int_0^{2\pi}\!\int_0^1 \sqrt{r^2+1}\sqrt{r^2+1}\, dr\, d\theta \\ &= \int_0^{2\pi}\!\int_0^1 r^2+1\, dr\, d\theta = \int_0^{2\pi} \frac{4}{3}\, d\theta = \frac{8\pi}{3}. \end{align*} \]
To finish up the problem, note that the surface area of the helicoid is \[ \int\!\!\int_S dS = \int_0^{2\pi}\!\int_0^1 \sqrt{r^2+1}\, dr\, d\theta. \]
Problem 2
2. Suppose \(\mathbf{F} = x\vi+y\vj+(z-2)\vk\). Calculate \(\displaystyle\int\!\!\int_S \mathbf{F}\cdot d\mathbf{S}\), for \(S\) the helicoid with parameterization \(G(u,v) = (u\cos(v), u\sin(v), v)\), with \(0\le u\le 1\) and \(0\le v\le 2\pi\). (5 points)
Solution. The parameterization is the same as the first problem, so we reuse the fact that \(\mathbf{T}_u\times\mathbf{T}_v = (\sin(v), -\cos(v), u)\). Doing so allows us to calculate
\[ \begin{align*} \int\!\!\int_S \mathbf{F}\cdot d\mathbf{S} &= \int_0^{2\pi}\!\int_0^1 \mathbf{F}(G(u,v))\cdot \mathbf{T}_u\times\mathbf{T}_v\, du\, dv = \int_0^{2\pi}\!\int_0^1 (u\cos v, u\sin v, v-2)\cdot(\sin v, -\cos v, u)\, du\, dv \\ &= \int_0^{2\pi}\!\int_0^1 uv-2u\, du\, dv = \int_0^{2\pi} \left(\frac{u^2v}{2}-u^2\right)\Bigg|_0^1 dv = \int_0^{2\pi} \frac{v}{2}-1\, dv = \left(\frac{v^2}{4}-v\right)\Bigg|_0^{2\pi} = \pi^2-2\pi. \end{align*} \]