2022-04-11

Find the surface area of the helicoid (spiral ramp) given by 𝑟⃗(𝑢, 𝑣) = 〈𝑢 cos 𝑣 , 𝑢 sin 𝑣 , 𝑣〉 for 0 ≤ 𝑢 ≤ 1 and 0 ≤ 𝑣 ≤ 𝜋. Use the following function to graph the helicoid in GeoGebra. surface(x,y,z, parameter variable 1, lower bound, upper bound, parameter variable 2, lower bound, upper bound) Vasquez

The helicoid is given by the vector equation:
$𝐫\left(u,v\right)=⟨u\mathrm{cos}v,u\mathrm{sin}v,v⟩\phantom{\rule{2em}{0ex}}0\le u\le 1,0\le v\le \pi$
To find its surface area, we need to compute the magnitude of the partial derivatives ${𝐫}_{u}$ and ${𝐫}_{v}$ and take their cross product:
${𝐫}_{u}=⟨\mathrm{cos}v,\mathrm{sin}v,0⟩$
${𝐫}_{v}=⟨-u\mathrm{sin}v,u\mathrm{cos}v,1⟩$
${𝐫}_{u}×{𝐫}_{v}=|\begin{array}{ccc}𝐢& 𝐣& 𝐤\\ \mathrm{cos}v& \mathrm{sin}v& 0\\ -u\mathrm{sin}v& u\mathrm{cos}v& 1\end{array}|=⟨-u\mathrm{sin}v,-u\mathrm{cos}v,u⟩$
The magnitude of this vector is $\sqrt{\left(u\mathrm{sin}v{\right)}^{2}+\left(u\mathrm{cos}v{\right)}^{2}+{u}^{2}}=u\sqrt{2}$.
Therefore, the surface area of the helicoid is given by the integral:
$A={\int }_{0}^{1}{\int }_{0}^{\pi }‖{𝐫}_{u}×{𝐫}_{v}‖dudv={\int }_{0}^{1}{\int }_{0}^{\pi }u\sqrt{2}dudv$
Evaluating this integral, we get:
$A=\sqrt{2}{\int }_{0}^{\pi }{\int }_{0}^{1}ududv=\frac{\pi }{\sqrt{2}}$
Therefore, the surface area of the helicoid is $\frac{\pi }{\sqrt{2}}$.

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