find the exact area of the surface obtained by rotating the curve about the x - axis.\n$y = \\sqrt{8 - x}$…

Explanation:

Step1: Find the derivative of (y)

Given (y = \sqrt{8 - x}=(8 - x)^{\frac{1}{2}}). Using the power - rule ((u^n)^\prime=nu^{n - 1}u^\prime), where (u = 8 - x), (n=\frac{1}{2}), and (u^\prime=-1). (y^\prime=\frac{1}{2}(8 - x)^{-\frac{1}{2}}\times(-1)=-\frac{1}{2\sqrt{8 - x}}). Then ((y^\prime)^2=\frac{1}{4(8 - x)}).

Step2: Use the surface - area formula

The formula for the surface area (S) of a surface obtained by rotating the curve (y = f(x)) about the (x) - axis from (a) to (b) is (S=\int_{a}^{b}2\pi y\sqrt{1+(y^\prime)^2}dx). Here, (a = 2), (b = 8), (y=\sqrt{8 - x}), and ((y^\prime)^2=\frac{1}{4(8 - x)}). So (1+(y^\prime)^2=1+\frac{1}{4(8 - x)}=\frac{4(8 - x)+1}{4(8 - x)}=\frac{32-4x + 1}{4(8 - x)}=\frac{33-4x}{4(8 - x)}). And (S=\int_{2}^{8}2\pi\sqrt{8 - x}\sqrt{\frac{33-4x}{4(8 - x)}}dx). Simplify the integrand: (\sqrt{8 - x}\sqrt{\frac{33-4x}{4(8 - x)}}=\frac{\sqrt{33-4x}}{2}). So (S=\int_{2}^{8}\pi\sqrt{33-4x}dx).

Step3: Use substitution

Let (u = 33-4x), then (du=-4dx). When (x = 2), (u = 33-4\times2=25); when (x = 8), (u = 33-4\times8=1). (dx=-\frac{1}{4}du). (S=\pi\int_{25}^{1}u^{\frac{1}{2}}\left(-\frac{1}{4}\right)du). Using the property (\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx), we have (S=\frac{\pi}{4}\int_{1}^{25}u^{\frac{1}{2}}du).

Step4: Integrate

Using the power - rule for integration (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)). (\frac{\pi}{4}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\big|{1}^{25}=\frac{\pi}{6}(u^{\frac{3}{2}})\big|{1}^{25}). Substitute the upper and lower limits: (\frac{\pi}{6}(25^{\frac{3}{2}}-1^{\frac{3}{2}})). Since (25^{\frac{3}{2}}=(25^{\frac{1}{2}})^3 = 125) and (1^{\frac{3}{2}}=1).

Answer:

(\frac{124\pi}{6}=\frac{62\pi}{3})