How do you find the surface area of the solid obtained by rotating about the x-axis the region bounded by 9x=y^2+18 on the interval 2<=x<=6 ?

Sep 20, 2014

The surface area can be found by

$S = 2 \pi {\int}_{0}^{6} y \sqrt{1 + \frac{4 {y}^{2}}{81}} \mathrm{dy} = 49 \pi$

Let us look at some details.

9x=y^2+18 Leftrightarrow x=y^2/9+2 Leftrightarrow y=pm3sqrt{x-2}

By taking the derivative,

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{2 x}{9}$

As x goes from 2 to 6, y goes from 0 to 6.

Since the surface area of revolution can be found by

$S = 2 \pi {\int}_{a}^{b} y \setminus \sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} \mathrm{dy}$,

we have

$S = 2 \pi {\int}_{0}^{6} y \sqrt{1 + \frac{4 {y}^{2}}{81}} \mathrm{dy}$

by substitution $u = 1 + \frac{4 {y}^{2}}{81}$. Rightarrow {du}/dy={8y}/81 Rightarrow dy=81/{8y}du,
As y goes from 0 to 6, u goes from 1 to 25/9.

$= 2 \pi {\int}_{1}^{\frac{25}{9}} y \sqrt{u} \frac{81}{8 y} \mathrm{du}$

by simplifying,

$= \frac{81 \pi}{4} {\int}_{1}^{\frac{25}{9}} {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{81 \pi}{4} {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{1}^{\frac{25}{9}}$

$= \frac{27 \pi}{2} \left[{\left(\frac{25}{9}\right)}^{\frac{3}{2}} - 1\right] = 49 \pi$