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

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

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Answer 1 · 2014-09-20T23:17:18

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

Let us look at some details.

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

By differentiating with respect to $y$ $y$ ,

$\frac{d x}{d y} = 4 y$ ${dx}/{dy}=4y$

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

$S = 2 π \int_{1}^{2} y \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$ $S=2pi int_1^2 ysqrt{1+({dx}/{dy})^2} dy$

$= 2 π \int_{1}^{2} y \sqrt{1 + 16 y^{2}} d y$ $=2pi int_1^2 ysqrt{1+16y^2}dy$

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

$y : 1 \to 2 \Rightarrow u : 17 \to 65$ $y: 1 to 2 Rightarrow u: 17 to 65$

$= 2 π \int_{17}^{65} y \sqrt{u} \frac{d u}{32 y}$ $=2pi int_{17}^{65}ysqrt{u}{du}/{32y}$

$= \frac{π}{16} \int_{17}^{65} u^{\frac{1}{2}} d u$ $=pi/16 int_{17}^{65}u^{1/2}du$

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

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

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

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