Find the exact area of the surface obtained by rotating the curve about the x-axis. y = √8 - x, 2 ≤x ≤8

Solution:

Given, y = √(8-x), 2 ≤x ≤8

We have to find the area of the surface by rotating the curve about the x-axis.

For rotation about the x-axis, the surface area formula is given by

\(S=2\pi \int_{a}^{b}y\sqrt{1+(y')^{2}}\, dx\)

y = √(8-x)

y’ = -1/2(√(8-x))

By rotating the curve y = √(8 - x) about the x-axis in the interval [2, 8]

\(S=2\pi \int_{2}^{8}\sqrt{8-x}\sqrt{1+(\frac{-1}{2\sqrt{8-x}})^{2}}\, dx\\S=2\pi \int_{2}^{8}\sqrt{8-x}\sqrt{1+\frac{1}{4(8-x)}}\, dx\)

\(S=2\pi \int_{2}^{8}\sqrt{8-x}\sqrt{1+\frac{1}{32-4x}}\, dx\\S=2\pi \int_{2}^{8}\sqrt{8-x}\sqrt{\frac{32-4x+1}{32-4x}}\, dx\)

\(S=2\pi \int_{2}^{8}\sqrt{8-x}\sqrt{\frac{33-4x}{32-4x}}\, dx\)

\(S=2\pi \int_{2}^{8}\sqrt{8-x}\frac{\sqrt{33-4x}}{\sqrt{4(8-x)}}\, dx\\S=2\pi \int_{2}^{8}\sqrt{8-x}\frac{\sqrt{33-4x}}{2\sqrt{(8-x)}}\, dx\\S=\pi \int_{2}^{8}\sqrt{8-x}\frac{\sqrt{33-4x}}{\sqrt{(8-x)}}\, dx\)

\(S=\pi \int_{2}^{8}\sqrt{33-4x}\, dx\)

Let u = 33 - 4x

du = -4 dx

So, dx = du/-4

Now, \(S=\pi \int_{2}^{8}\sqrt{u}(\frac{du}{-4})\\S=\frac{\pi }{-4}\int_{2}^{8}\sqrt{u}\, du\)

\(S=\frac{\pi }{-4}[\frac{2u^{\frac{3}{2}}}{3}]_{2}^{8}\)

\(S=\frac{2\pi }{-4(3)}[(33-4x)^{\frac{3}{2}}]_{2}^{8}\\S=\frac{\pi }{-6}[(33-4x)^{\frac{3}{2}}]_{2}^{8}\)

\(S=\frac{\pi }{-6}[(33-4(8))^{\frac{3}{2}}-(33-4(2))^{\frac{3}{2}}]\)

\(S=\frac{\pi }{-6}[(33-32)^{\frac{3}{2}}-(33-8)^{\frac{3}{2}}]\)

\(S=\frac{\pi }{-6}[(1)^{\frac{3}{2}}-(25)^{\frac{3}{2}}]\)

\(S=\frac{\pi }{-6}[1-125]\)

\(S=\frac{\pi }{-6}[-124]\\S=\pi (20.67)\)

S = 20.67(3.14)

S = 64.89 square units

Therefore, the exact area of the surface is 64.89 square units.

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Find the exact area of the surface obtained by rotating the curve about the x-axis. y = √8 - x, 2 ≤ x ≤8

Summary:

The exact area of the surface obtained by rotating the curve about the x-axis. y = √8 - x, 2 ≤ x ≤8 is 64.89 square units.