estimate the area of the surface generated by revolving the curve y = cos(2x),0 ≤ x ≤ π/4 about the x…

Explanation:

Step1: Recall surface - area formula

The formula for the surface area (S) of the surface generated by revolving the curve (y = f(x)) about the (x) - axis from (x=a) to (x = b) is (S=2\pi\int_{a}^{b}y\sqrt{1+(y')^{2}}dx). First, find (y'): If (y=\cos(2x)), then (y'=- 2\sin(2x)). So, (1+(y')^{2}=1 + 4\sin^{2}(2x)), and the surface - area integral is (S = 2\pi\int_{0}^{\frac{\pi}{4}}\cos(2x)\sqrt{1 + 4\sin^{2}(2x)}dx).

Step2: Set up the trapezoidal rule

The trapezoidal rule for (\int_{a}^{b}g(x)dx) with (n) sub - divisions is (\int_{a}^{b}g(x)dx\approx\frac{\Delta x}{2}[g(x_0)+2g(x_1)+2g(x_2)+\cdots+2g(x_{n - 1})+g(x_n)]), where (\Delta x=\frac{b - a}{n}). Here, (a = 0), (b=\frac{\pi}{4}), (n = 4), so (\Delta x=\frac{\frac{\pi}{4}-0}{4}=\frac{\pi}{16}). The (x) - values are (x_0 = 0), (x_1=\frac{\pi}{16}), (x_2=\frac{\pi}{8}), (x_3=\frac{3\pi}{16}), (x_4=\frac{\pi}{4}). Let (g(x)=\cos(2x)\sqrt{1 + 4\sin^{2}(2x)}).

Step3: Calculate (g(x_i)) for (i = 0,1,2,3,4)

• When (x_0 = 0): (g(x_0)=\cos(0)\sqrt{1+4\sin^{2}(0)}=1\times\sqrt{1 + 0}=1).
• When (x_1=\frac{\pi}{16}): (\sin(2x_1)=\sin(\frac{\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{2}), (\cos(2x_1)=\cos(\frac{\pi}{8})=\frac{\sqrt{2 + \sqrt{2}}}{2}). (g(x_1)=\cos(\frac{\pi}{8})\sqrt{1 + 4\sin^{2}(\frac{\pi}{8})}=\frac{\sqrt{2+\sqrt{2}}}{2}\sqrt{1+4\times\frac{2 - \sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}\sqrt{3-\sqrt{2}}).
• When (x_2=\frac{\pi}{8}): (\sin(2x_2)=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}), (\cos(2x_2)=\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}). (g(x_2)=\cos(\frac{\pi}{4})\sqrt{1 + 4\sin^{2}(\frac{\pi}{4})}=\frac{\sqrt{2}}{2}\sqrt{1 + 4\times\frac{1}{2}}=\frac{\sqrt{2}}{2}\sqrt{3}).
• When (x_3=\frac{3\pi}{16}): (\sin(2x_3)=\sin(\frac{3\pi}{8})=\frac{\sqrt{2+\sqrt{2}}}{2}), (\cos(2x_3)=\cos(\frac{3\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{2}). (g(x_3)=\cos(\frac{3\pi}{8})\sqrt{1 + 4\sin^{2}(\frac{3\pi}{8})}=\frac{\sqrt{2-\sqrt{2}}}{2}\sqrt{1+4\times\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2-\sqrt{2}}}{2}\sqrt{3+\sqrt{2}}).
• When (x_4=\frac{\pi}{4}): (\sin(2x_4)=\sin(\frac{\pi}{2}) = 1), (\cos(2x_4)=\cos(\frac{\pi}{2})=0). (g(x_4)=\cos(\frac{\pi}{2})\sqrt{1 + 4\sin^{2}(\frac{\pi}{2})}=0).

Step4: Apply the trapezoidal rule

(\int_{0}^{\frac{\pi}{4}}g(x)dx\approx\frac{\frac{\pi}{16}}{2}[g(x_0)+2g(x_1)+2g(x_2)+2g(x_3)+g(x_4)]) (=\frac{\pi}{32}[1 + 2\times\frac{\sqrt{2+\sqrt{2}}}{2}\sqrt{3-\sqrt{2}}+2\times\frac{\sqrt{2}}{2}\sqrt{3}+2\times\frac{\sqrt{2-\sqrt{2}}}{2}\sqrt{3+\sqrt{2}}+0]). (S\approx2\pi\times\frac{\pi}{32}[1+\sqrt{2+\sqrt{2}}\sqrt{3-\sqrt{2}}+\sqrt{2}\sqrt{3}+\sqrt{2-\sqrt{2}}\sqrt{3+\sqrt{2}}]). Using a calculator: [ \begin{align*} S&\approx2\pi\times\frac{\pi}{32}(1 + 2.0976+1.9487+2.0976)\ &=\frac{\pi^{2}}{16}(7.1439)\ &\approx4.3902 \end{align*} ]

Answer:

(4.3902)