problem 2: surface area of revolution\nprovide the general set - up to find the surface area of the surface…

Explanation:

Step1: Recall surface - area formula

The formula for the surface area $S$ of the surface obtained by rotating the curve $y = f(x)$ for $a\leq x\leq b$ about the $x -$axis is $S=\int_{a}^{b}2\pi y\sqrt{1+(y')^{2}}dx$.

Step2: Find the derivative of $y$

Given $y = 2\sqrt{x + 1}$, using the power - rule, if $y=2(x + 1)^{\frac{1}{2}}$, then $y'=\frac{2}{2\sqrt{x + 1}}=\frac{1}{\sqrt{x + 1}}$.

Step3: Substitute $y$ and $y'$ into the surface - area formula

We have $a = 2$, $b = 7$, $y = 2\sqrt{x + 1}$, and $y'=\frac{1}{\sqrt{x + 1}}$. Then $1+(y')^{2}=1+\frac{1}{x + 1}=\frac{x + 1+1}{x + 1}=\frac{x + 2}{x + 1}$. And $\sqrt{1+(y')^{2}}=\sqrt{\frac{x + 2}{x + 1}}$. Also, $y = 2\sqrt{x + 1}$. So the surface - area formula becomes $S=\int_{2}^{7}2\pi(2\sqrt{x + 1})\sqrt{\frac{x + 2}{x + 1}}dx$. Simplify the integrand: $(2\sqrt{x + 1})\sqrt{\frac{x + 2}{x + 1}}=2\sqrt{x + 2}$. So $S = 4\pi\int_{2}^{7}\sqrt{x + 2}dx$.

Step4: Integrate

Let $u=x + 2$, then $du=dx$. When $x = 2$, $u = 4$; when $x = 7$, $u = 9$. The integral becomes $4\pi\int_{4}^{9}u^{\frac{1}{2}}du$. Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $4\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{4}^{9}$.

Step5: Evaluate the definite integral

$4\pi\times\frac{2}{3}(9^{\frac{3}{2}}-4^{\frac{3}{2}})=4\pi\times\frac{2}{3}(27 - 8)=\frac{8\pi}{3}\times19=\frac{152\pi}{3}$.

The general set - up for the surface area of the surface obtained by rotating the curve $y = f(x)$, $a\leq x\leq b$ about the $x -$axis is $S=\int_{a}^{b}2\pi f(x)\sqrt{1+(f'(x))^{2}}dx$.

Answer:

The general set - up is $S=\int_{a}^{b}2\pi f(x)\sqrt{1+(f'(x))^{2}}dx$. The exact surface area for $y = 2\sqrt{x + 1}$, $2\leq x\leq 7$ is $\frac{152\pi}{3}$.