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 $1<=y<=2$ ?

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Answer 1 · 2014-09-20T23:17:18

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

Let us look at some details.

$x=1+2y^2$

By differentiating with respect to $y$ ,

${dx}/{dy}=4y$

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

$S=2pi int_1^2 ysqrt{1+({dx}/{dy})^2} dy$

$=2pi int_1^2 ysqrt{1+16y^2}dy$

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

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

$=2pi int_{17}^{65}ysqrt{u}{du}/{32y}$

$=pi/16 int_{17}^{65}u^{1/2}du$

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

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

How do you find the surface area of the solid obtained by rotating about the xx-axis the region bounded by x=1+2y^2x=1+2y^2 on the interval 1<=y<=21<=y<=2 ?