**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 \(\displaystyle \sqrt{(dy)^2+(dx)^2}\) rather than dx from x=3 to 11,

having y as radius.

\(\displaystyle \sqrt{(dy)^2+(dx)^2}=\sqrt{\left(\frac{dy}{dx}\right)^2+\left(\frac{dx}{dx}\right)^2}\ dx\)

\(\displaystyle f(x)=(2+2x)^{0.5}\)

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

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

\(\displaystyle =\sqrt{\frac{2+2x+1}{2+2x}}\)

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

\(\displaystyle =2{\pi}\int_{3}^{11}{\sqrt{3+2x}}\ dx\)

\(\displaystyle u=3+2x\)

\(\displaystyle du=2dx\)

\(\displaystyle dx=\frac{du}{2}\)

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.

\(\displaystyle {\pi}\int_{9}^{25}{u^{\frac{1}{2}}}\ du\)

Or

\(\displaystyle {\pi}\int_{x=3}^{x=11}{u^{\frac{1}{2}}}\ du\)

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

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