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

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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 $9 x = y^{2} + 18$ $9x=y^2+18$ on the interval $2 \leq x \leq 6$ $2<=x<=6$ ?

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Answer 1 · 2014-09-20T13:34:26

The surface area can be found by

$S = 2 π \int_{0}^{6} y \sqrt{1 + \frac{4 y^{2}}{81}} d y = 49 π$ $S=2piint_0^6 ysqrt{1+{4y^2}/81}dy=49pi$

Let us look at some details.

$9 x = y^{2} + 18 \Leftrightarrow x = \frac{y^{2}}{9} + 2 \Leftrightarrow y = \pm 3 \sqrt{x - 2}$ $9x=y^2+18 Leftrightarrow x=y^2/9+2 Leftrightarrow y=pm3sqrt{x-2}$

By taking the derivative,

$\frac{d x}{d y} = \frac{2 x}{9}$ $dx/dy={2x}/9$

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

Since the surface area of revolution can be found by

$S = 2 π \int_{a}^{b} y \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$ $S=2piint_a^b y\sqrt{1+(dx/dy)^2}dy$ ,

we have

$S = 2 π \int_{0}^{6} y \sqrt{1 + \frac{4 y^{2}}{81}} d y$ $S=2piint_0^6 ysqrt{1+{4y^2}/81}dy$

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

$= 2 π \int_{1}^{\frac{25}{9}} y \sqrt{u} \frac{81}{8 y} d u$ $=2pi int_1^{25/9}ysqrt{u}{81}/{8y}du$

by simplifying,

$= \frac{81 π}{4} \int_{1}^{\frac{25}{9}} u^{\frac{1}{2}} d u$ $={81pi}/4 int_1^{25/9}u^{1/2}du$

$= \frac{81 π}{4} {[\frac{2}{3} u^{\frac{3}{2}}]}_{1}^{\frac{3}{2}}$ $={81pi}/4[2/3u^{3/2}]_1^{25/9}$

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

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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 $9 x = y^{2} + 18$ $9x=y^2+18$ on the interval $2 \leq x \leq 6$ $2<=x<=6$ ?

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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 $9 x = y^{2} + 18$ $9x=y^2+18$ on the interval $2 \leq x \leq 6$ $2<=x<=6$ ?