1. ## surface area about x-axis

I need to find the surface area of the above curve, about the x-axis. I'm having trouble taking the integral. Once everthing is set up its:

$\displaystyle 2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)^2))}$ from 4 to 9

Or at least, that's what I think it is..
Any help would be greatly appreciated.

2. Originally Posted by cdlegendary

I need to find the surface area of the above curve, about the x-axis. I'm having trouble taking the integral. Once everthing is set up its:

$\displaystyle 2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)^2))}$ from 4 to 9

Or at least, that's what I think it is..
Any help would be greatly appreciated.
As you know, the surface area of a revolution is found through the following formula:

$\displaystyle A = 2\pi \int _a ^b y \sqrt{1+(\frac{dy}{dx})^2} dx$

Your integral is correct except for the $\displaystyle (\frac{dy}{dx})^2$ term, just need to recheck the square.

3. Ah, I made a typo. so the integral would be:

$\displaystyle 2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)))}$

since the derivative of $\displaystyle \sqrt(x)$ is $\displaystyle 1/(2\sqrt(x))$ is it not?

4. Originally Posted by cdlegendary
Ah, I made a typo. so the integral would be:

$\displaystyle 2pi\int(\sqrt{x})(\sqrt{(1+(1/4x)))}$

since the derivative of $\displaystyle \sqrt(x)$ is $\displaystyle 1/(2\sqrt(x))$ is it not?
You bet.

5. thanks, I'm still having trouble taking the integral though. any pointers?

6. How about that $\displaystyle \sqrt{x}\sqrt{1+\frac1{4x}}=\sqrt{x\left(1+\frac1{ 4x}\right)}=\sqrt{x+\frac14}=\sqrt{\frac{4x+1}4}=\ frac{\sqrt{4x+1}}2$?

--Kevin C.