find the area of the surface obtained by rotating the curve $y = \\sin(4x)$ about the $x$-axis from $x = 0$…

Explanation:

Step1: Recall surface - area formula

The formula for the surface area $S$ of the surface obtained by rotating 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 the derivative of $y=\sin(4x)$. Using the chain - rule, if $y = \sin(u)$ and $u = 4x$, then $y'=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{dy}{du}=\cos(u)$ and $\frac{du}{dx}=4$, we have $y' = 4\cos(4x)$.

Step2: Substitute $y$ and $y'$ into the formula

Substitute $y=\sin(4x)$ and $y' = 4\cos(4x)$ into the surface - area formula. We get $S = 2\pi\int_{0}^{\frac{\pi}{4}}\sin(4x)\sqrt{1 + 16\cos^{2}(4x)}dx$. Let $u=\cos(4x)$, then $du=-4\sin(4x)dx$. When $x = 0$, $u=\cos(0)=1$; when $x=\frac{\pi}{4}$, $u=\cos(\pi)=-1$. The integral becomes $S=2\pi\int_{1}^{-1}-\frac{1}{4}\sqrt{1 + 16u^{2}}du=\frac{\pi}{2}\int_{-1}^{1}\sqrt{1 + 16u^{2}}du$.

Step3: Use the integral formula for $\sqrt{a^{2}+x^{2}}$

The integral formula for $\int\sqrt{a^{2}+x^{2}}dx=\frac{x}{2}\sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2}\ln(x+\sqrt{a^{2}+x^{2}})+C$. Here $a = \frac{1}{4}$ and $x = u$. So $\int\sqrt{1 + 16u^{2}}du=\int\sqrt{1+(4u)^{2}}du$. Let $t = 4u$, $dt=4du$, then $\int\sqrt{1+(4u)^{2}}du=\frac{1}{4}\int\sqrt{1 + t^{2}}dt=\frac{1}{4}(\frac{t}{2}\sqrt{1 + t^{2}}+\frac{1}{2}\ln(t+\sqrt{1 + t^{2}}))+C=\frac{u}{2}\sqrt{1 + 16u^{2}}+\frac{1}{8}\ln(4u+\sqrt{1 + 16u^{2}})+C$.

Step4: Evaluate the definite integral

$S=\frac{\pi}{2}\left[\frac{u}{2}\sqrt{1 + 16u^{2}}+\frac{1}{8}\ln(4u+\sqrt{1 + 16u^{2}})\right]_{-1}^{1}$. [ \begin{align*} S&=\frac{\pi}{2}\left[\left(\frac{1}{2}\sqrt{1 + 16}+\frac{1}{8}\ln(4+\sqrt{1 + 16})\right)-\left(-\frac{1}{2}\sqrt{1 + 16}+\frac{1}{8}\ln(-4+\sqrt{1 + 16})\right)\right]\ &=\frac{\pi}{2}\left[\sqrt{17}+\frac{1}{8}\left(\ln(4 + \sqrt{17})-\ln(\sqrt{17}-4)\right)\right]\ &=\frac{\pi}{2}\left[\sqrt{17}+\frac{1}{8}\ln\left(\frac{4+\sqrt{17}}{\sqrt{17}-4}\right)\right]\ &=\frac{\pi}{2}\left[\sqrt{17}+\frac{1}{8}\ln\left(\frac{(4 + \sqrt{17})^{2}}{17-16}\right)\right]\ &=\frac{\pi}{2}\left[\sqrt{17}+\frac{1}{4}\ln(4+\sqrt{17})\right] \end{align*} ]

Answer:

$\frac{\pi}{2}\left[\sqrt{17}+\frac{1}{4}\ln(4 + \sqrt{17})\right]$