area of a surface of revolution

**break** · May 21st 2010, 02:36 PM

Find the area of the surface obtained by rotating the curve about the x-axis.?

y = sqrt(2 + 2 x)

3<=x<=11

I know that the formula used for this is S =(integral from 3to11) 2*pi*y*sqrt(1+(y')^2)dy

for y' i get y'=1/( 2*sqrt(2+2x) )

once i plug in both y and y'. i get stuff because i'm not sure how it will all work out with the u-substitution and all.

can someone help? thanks!

**Archie Meade** · May 21st 2010, 03:44 PM

Originally Posted by break

Find the area of the surface obtained by rotating the curve about the x-axis.?

y = sqrt(2 + 2 x)

3<=x<=11

I know that the formula used for this is S =(integral from 3to11) 2*pi*y*sqrt(1+(y')^2)dy

for y' i get y'=1/( 2*sqrt(2+2x) )

once i plug in both y and y'. i get stuff because i'm not sure how it will all work out with the u-substitution and all.

can someone help? thanks!

The surface area is the integral of circle circumferences, multiplied by $\sqrt{(dy)^2+(dx)^2}$ rather than dx from x=3 to 11,
having y as radius.

$\sqrt{(dy)^2+(dx)^2}=\sqrt{\left(\frac{dy}{dx}\rig ht)^2+\left(\frac{dx}{dx}\right)^2}\ dx$

$f(x)=(2+2x)^{0.5}$

$\frac{d}{dx}\ f(x)=0.5(2+2x)^{-0.5}2=\frac{1}{\sqrt{2+2x}}$

$\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\fra c{1}{2+2x}}$

$=\sqrt{\frac{2+2x+1}{2+2x}}$

$\int_{3}^{11}{(2{\pi}r)}\sqrt{(dy)^2+(dx)^2}=2{\pi }\int_{3}^{11}{\sqrt{2+2x}}\sqrt{\frac{3+2x}{2+2x} }dx$

$=2{\pi}\int_{3}^{11}{\sqrt{3+2x}}\ dx$

$u=3+2x$

$du=2dx$

$dx=\frac{du}{2}$

If you like you can change the limits and work exclusively with u,
or leave the limits alone and switch back to x having evaluated the integral.

${\pi}\int_{9}^{25}{u^{\frac{1}{2}}}\ du$

Or

${\pi}\int_{x=3}^{x=11}{u^{\frac{1}{2}}}\ du$

$={\pi}\frac{2}{3}\left[(3+2x)^{\frac{3}{2}}\right]$ from x=3 to x=11

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

**drumist** · May 21st 2010, 03:59 PM

Originally Posted by Archie Meade

The surface area is the integral of circle circumferences, from x=3 to 11,
having y as radius.

$\int_{3}^{11}{(2{\pi}r)}dx=2{\pi}\int_{3}^{11}{\sq rt{2+2x}}\ dx$

This is incorrect! Please see this: Surface of Revolution -- from Wolfram MathWorld

**drumist** · May 21st 2010, 04:03 PM

Originally Posted by break

Find the area of the surface obtained by rotating the curve about the x-axis.?

y = sqrt(2 + 2 x)

3<=x<=11

I know that the formula used for this is S =(integral from 3to11) 2*pi*y*sqrt(1+(y')^2)dy

for y' i get y'=1/( 2*sqrt(2+2x) )

once i plug in both y and y'. i get stuff because i'm not sure how it will all work out with the u-substitution and all.

can someone help? thanks!

$y=\sqrt{2+2x} = (2+2x)^{1/2}$

$y' = \frac{1}{2} (2+2x)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2+2x}}$

So:

$1+(y')^2 = 1+ \frac{1}{2+2x} = \frac{3+2x}{2+2x}$

So:

$y \cdot \sqrt{1+(y')^2} = \sqrt{2+2x} \cdot \sqrt{\frac{3+2x}{2+2x}} = \sqrt{3+2x}$

So the integral to solve is

$2 \pi \int_3^{11} \sqrt{3+2x} \, dx$

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