x = 2\\cos(\\theta)-\\cos(2\\theta), y = 2\\sin(\\theta)-\\sin(2\\theta)\nif this curve is rotated about the…

Explanation:

Step1: Recall the surface - area formula for a curve rotated about the x - axis

The formula for the surface area (S) of a curve given by (x = x(\theta)) and (y = y(\theta)) rotated about the (x) - axis is (S=\int_{a}^{b}2\pi y\sqrt{(\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2}d\theta). First, find (\frac{dx}{d\theta}) and (\frac{dy}{d\theta}). Given (x = 2\cos\theta-\cos(2\theta)), then (\frac{dx}{d\theta}=- 2\sin\theta + 2\sin(2\theta)) by the chain - rule ((\frac{d}{d\theta}\cos(u)=-\sin(u)\frac{du}{d\theta})). Given (y = 2\sin\theta-\sin(2\theta)), then (\frac{dy}{d\theta}=2\cos\theta - 2\cos(2\theta)).

Step2: Calculate ((\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2)

[ \begin{align*} (\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2&=( - 2\sin\theta + 2\sin(2\theta))^2+(2\cos\theta - 2\cos(2\theta))^2\ &=4\sin^{2}\theta-8\sin\theta\sin(2\theta)+4\sin^{2}(2\theta)+4\cos^{2}\theta - 8\cos\theta\cos(2\theta)+4\cos^{2}(2\theta)\ &=4(\sin^{2}\theta+\cos^{2}\theta)+4(\sin^{2}(2\theta)+\cos^{2}(2\theta))-8(\cos\theta\cos(2\theta)+\sin\theta\sin(2\theta))\ &=4 + 4-8\cos(2\theta-\theta)\ &=8 - 8\cos\theta \end{align*} ] using the identities (\sin^{2}u+\cos^{2}u = 1) and (\cos(A - B)=\cos A\cos B+\sin A\sin B).

Step3: Determine the parameter interval

By observing the graph of the polar - like curve (x = 2\cos\theta-\cos(2\theta)), (y = 2\sin\theta-\sin(2\theta)), the curve is traced out once for (\theta\in[0,2\pi]).

Step4: Substitute into the surface - area formula

[ \begin{align*} S&=\int_{0}^{2\pi}2\pi(2\sin\theta-\sin(2\theta))\sqrt{8 - 8\cos\theta}d\theta\ &=2\pi\int_{0}^{2\pi}(2\sin\theta-\sin(2\theta))\sqrt{8(1 - \cos\theta)}d\theta\ &=4\sqrt{2}\pi\int_{0}^{2\pi}(2\sin\theta - 2\sin\theta\cos\theta)\sqrt{1 - \cos\theta}d\theta \end{align*} ] Let (u = 1-\cos\theta), then (du=\sin\theta d\theta). When (\theta = 0), (u = 0); when (\theta=2\pi), (u = 0). [ \begin{align*} S&=4\sqrt{2}\pi\int_{0}^{0}(2(1 - u^{\frac{1}{2}})-2(1 - u^{\frac{1}{2}})(1 - u))u^{\frac{1}{2}}du = 0 \end{align*} ] A better way is to use the double - angle formula (1-\cos\theta = 2\sin^{2}\frac{\theta}{2}). Then (\sqrt{8 - 8\cos\theta}=\sqrt{16\sin^{2}\frac{\theta}{2}} = 4|\sin\frac{\theta}{2}|). [ \begin{align*} S&=\int_{0}^{2\pi}2\pi(2\sin\theta-\sin(2\theta))\cdot4\sin\frac{\theta}{2}d\theta\ &=8\pi\int_{0}^{2\pi}(2\sin\theta-\sin(2\theta))\sin\frac{\theta}{2}d\theta\ &=8\pi\int_{0}^{2\pi}(2\cdot2\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2\sin\theta\cos\theta)\sin\frac{\theta}{2}d\theta\ &=16\pi\int_{0}^{2\pi}\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}d\theta-16\pi\int_{0}^{2\pi}\sin\theta\cos\theta\sin\frac{\theta}{2}d\theta \end{align*} ] For (\int_{0}^{2\pi}\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}d\theta), let (t=\sin\frac{\theta}{2}), (dt=\frac{1}{2}\cos\frac{\theta}{2}d\theta). When (\theta = 0), (t = 0); when (\theta = 2\pi), (t = 0). (\int_{0}^{2\pi}\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}d\theta = 2\int_{0}^{0}t^{2}dt = 0). For (\int_{0}^{2\pi}\sin\theta\cos\theta\sin\frac{\theta}{2}d\theta), use (\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}), then (\sin\theta\cos\theta\sin\frac{\theta}{2}=2\sin^{2}\frac{\theta}{2}\cos\frac{\theta}{2}\cos\theta). The correct way: [ \begin{align*} S&=\int_{0}^{2\pi}2\pi(2\sin\theta-\sin(2\theta))\sqrt{8 - 8\cos\theta}d\theta\ &=8\sqrt{2}\pi\int_{0}^{2\pi}(2\sin\theta-\sin(2\theta))\sqrt{1 - \cos\theta}d\theta\ &=8\sqrt{2}\pi\int_{0}^{2\pi}(2\sin\theta - 2\sin\theta\cos\theta)\sqrt{1 - \cos\theta}d\theta\ &=16\sqrt{2}\pi\int_{0}^{2\pi}\sin\theta(1 - \cos\theta)\sqrt{1 - \cos\theta}d\theta \end{align*} ] Let (u = 1-\cos\theta), (du=\sin\theta d\theta). When (\theta = 0), (u = 0); when (\theta = 2\pi), (u = 0). [ \begin{align*} S&=16\sqrt{2}\pi\int_{0}^{2}u^{\frac{3}{2}}du\ &=16\sqrt{2}\pi\cdot\frac{2}{5}u^{\frac{5}{2}}\big|_{0}^{2}\ &=\frac{64\pi}{5}\sqrt{2}\cdot2^{\frac{1}{2}}\ &=\frac{128\pi}{5} \end{align*} ]

Answer:

(\frac{128\pi}{5})