need help doing an integral by hand to find surface area

[studentMCCS]

I can't figure out how to do this integral by hand.

I am finding the surface area of the the function y=(x+1)^(1/2) from [1,5], when revolved around the x-axis.

dy/dx=1/2(x+1)^(1/2)

(dy/dx)^2=1/(4(x+1))

Surface Area = 2PI * integral (((x+1)^(1/2))*(1+(1/4(x+1))^(1/2) , x, 1, 5)

 

[tkhunny]

You may wish to start with the correct derivative.

\(\displaystyle \frac{d}{dx}(1+x)^{\frac{1}{2}} = \frac{1}{2\cdot (1+x)^{\frac{1}{2}}}\)

Somehoe, yours didn't end up with anything in the denominator.

 

[studentMCCS]

Never mind, I figured it out.

I just had to transform into

((x+1)(1+(1/(4(x+1))))^(1/2)
=(x+(5/4))^(1/2)
u=x+5/4
du=dx
...
Surface Area=(49PI)/3

 

[studentMCCS]

You may wish to start with the correct derivative.

\(\displaystyle \frac{d}{dx}(1+x)^{\frac{1}{2}} = \frac{1}{2\cdot (1+x)^{\frac{1}{2}}}\)

Somehoe, yours didn't end up with anything in the denominator.

Sorry, I wrote that out wrong. I meant 1/(2(x+1))^(1/2))

Thanks for pointing that out.

I'm going to have to learn how to use \(\displaystyle .\)