How do you find the surface area of the solid obtained by rotating about the x-axis the region bounded by x=1+2y^2 on the interval 1<=y<=2 ?

Sep 20, 2014

The surface area is $\frac{\pi}{24} \left[{\left(65\right)}^{\frac{3}{2}} - {\left(17\right)}^{\frac{3}{2}}\right]$.

Let us look at some details.

$x = 1 + 2 {y}^{2}$

By differentiating with respect to $y$,

$\frac{\mathrm{dx}}{\mathrm{dy}} = 4 y$

So, we can find the surface area $S$ by

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

$= 2 \pi {\int}_{1}^{2} y \sqrt{1 + 16 {y}^{2}} \mathrm{dy}$

by substitution $u = 1 + 16 {y}^{2}$.
Rightarrow {du}/{dy}=32y Rightarrow{dy}/{du}=1/{32y} Rightarrow dy={du}/{32y}

$y : 1 \to 2 R i g h t a r r o w u : 17 \to 65$

$= 2 \pi {\int}_{17}^{65} y \sqrt{u} \frac{\mathrm{du}}{32 y}$

$= \frac{\pi}{16} {\int}_{17}^{65} {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{\pi}{16} {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{17}^{65}$

$= \frac{\pi}{24} \left[{\left(65\right)}^{\frac{3}{2}} - {\left(17\right)}^{\frac{3}{2}}\right]$