# Thread: Tough integral (for me)

1. ## Tough integral (for me)

The initial question was (if anyone is interested):
Find the area of the surface obtained by rotating the curve

but I am stuck at the integral mentioned above. Having just checked. I do not get the right answer when I use math software so I guess I did something wrong prior to getting this integral.

Here is the integral I am finding tough to do:
integrate 4&#x2a;pi&#x2a;sqrt&#x28;x&#x29;&#x2a;sqrt&#x28;&# x28;sqrt&#x28;x&#x29; &#x2b; 2&#x29;&#x2f;sqrt&#x28;x&#x29;&#x29;, from 0 to 2 - Wolfram|Alpha

I am probably not using the right integral since using math software gives me the wrong answer so any help would be greatly appreciated!

2. Originally Posted by s3a
The initial question was (if anyone is interested):

Find the area of the surface obtained by rotating the curve

but I am stuck at the integral mentioned above. Having just checked. I do not get the right answer when I use math software so I guess I did something wrong prior to getting this integral.

Here is the integral I am finding tough to do:
integrate 4&#x2a;pi&#x2a;sqrt&#x28;x&#x29;&#x2a;sqrt&#x28;&# x28;sqrt&#x28;x&#x29; &#x2b; 2&#x29;&#x2f;sqrt&#x28;x&#x29;&#x29;, from 0 to 2 - Wolfram|Alpha

I am probably not using the right integral since using math software gives me the wrong answer so any help would be greatly appreciated!

$S=2\pi\int_a^bf(x)\sqrt{1+(f'(x))^2}\,dx$
$f(x)=\sqrt{4x}=2x^{\frac{1}{2}}$
$f'(x)=x^{-\frac{1}{2}}$
$S=2\pi\int_0^2 2\sqrt{x}\sqrt{1+\frac{1}{x}}\,dx$