# How do you find the exact area of the surface obtained rotating the curve about the x-axis of y=sqrt(8-x), 2<=x<=8?

Jun 27, 2018

graph{y = sqrt(8-x) [-1.07, 10.97, -1.25, 5.76]}

$\mathrm{dS} = 2 \pi y \mathrm{ds}$

Arc length is:

$\mathrm{ds} = \sqrt{1 + x {'}^{2}} \mathrm{dy}$

$y = \sqrt{8 - x} q \quad x = 8 - {y}^{2} q \quad x ' = - 2 y$

$x : 2 \rightarrow 8 q \quad \leftrightarrow q \quad y : \sqrt{6} \rightarrow 0$

$S = 2 \pi {\int}_{C} y \sqrt{1 + x {'}^{2}} \mathrm{dy}$

$= 2 \pi {\int}_{0}^{\sqrt{6}} y \sqrt{1 + 4 {y}^{2}} \mathrm{dy}$

$= 2 \pi {\int}_{0}^{\sqrt{6}} \frac{d}{\mathrm{dy}} \left(\frac{1}{12} {\left(1 + 4 {y}^{2}\right)}^{\frac{3}{2}}\right) \mathrm{dy}$

$= \frac{\pi}{6} {\left[{\left(1 + 4 {y}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{\sqrt{6}}$

$= \frac{\pi}{6} \left(\left[{\left(1 + 24\right)}^{\frac{3}{2}}\right] - \left[{\left(1 + 0\right)}^{\frac{3}{2}}\right]\right)$

$= \frac{62 \pi}{3}$