# Integrate sqrt(1+(1\sqrt(x)))^2

• Jul 15th 2010, 06:05 AM
MechEng
Integrate sqrt(1+(1\sqrt(x)))^2
Good morning,

I left off on this problem yesterday. I am given:

Find the surface area by revolving around the x-axis:
$\displaystyle f(x) = 2\sqrt{x}$ for $\displaystyle 1\leq x \leq4$

So far I have the following:

$\displaystyle f'(x) = \frac{1}{\sqrt{x}}$

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

So...

$\displaystyle L = \int_{1}^{4}{\sqrt{\frac{x+1}{x}} dx$

And... I'm stuck.

I don't now why roots are giving me so much trouble lately. Is there something that I have forgotten or seem to be missing? I don't recall roots being this problematic.
• Jul 15th 2010, 06:22 AM
Ted
The problems wants you to find the surface area not the length of the curve.
• Jul 15th 2010, 07:01 AM
MechEng
Yes, it does want me to find the surface area. However, I still need to be able to integrate $\displaystyle \int_{1}^{4}{\sqrt{\frac{x+1}{x}} dx$ as $\displaystyle S=\int_{1}^{4}{2\pi\sqrt{\frac{x+1}{x}} dx$ is the same as $\displaystyle S=2\pi\int_{1}^{4}{\sqrt{\frac{x+1}{x}} dx$ ... No?

Is finding the surface not as simple as multiplying the lentgh of the curve by $\displaystyle 2\pi$?
• Jul 15th 2010, 08:25 AM
Ted
Quote:

Originally Posted by MechEng
Is finding the surface not as simple as multiplying the lentgh of the curve by $\displaystyle 2\pi$?

Becareful, you are multiplying the dS (which is $\displaystyle \sqrt{1+f^2(x)}$) not the length of the curve!

For your question, you will multiply the dS by $\displaystyle 2\pi y$ if the revolution about the x-axis.
and you will multiply it by $\displaystyle 2\pi x$ if the revolution about the y-axis.

So the required surface area is:

$\displaystyle 4\pi \int_1^4 \sqrt{x} \, \sqrt{\frac{x+1}{x}} \, dx$

Can you solve it?
• Jul 15th 2010, 09:14 AM
MechEng
I think so...

Jumping to the end, I get:

$\displaystyle S = \frac{8\pi(5\sqrt{5}-2\sqrt{2})}{3}$

Does that look correct to you?
• Jul 15th 2010, 09:38 AM
Ted
That 8 should be 4.
• Jul 15th 2010, 10:01 AM
MechEng
$\displaystyle S=\int_{1}^{4}{2\pi2\sqrt{x}\sqrt{\frac{x+1}{x}}} dx$

$\displaystyle S=4\pi\int_{1}^{4}{\sqrt{x}\sqrt{\frac{x+1}{x}}} dx$

$\displaystyle S=4\pi\int_{1}^{4}{x^{\frac{1}{2}}(x+1)^{\frac{1}{ 2}}x^{-\frac{1}{2}}} dx$

$\displaystyle S=4\pi\int_{1}^{4}{(x+1)^{\frac{1}{2}}} dx$
• Jul 15th 2010, 10:09 AM
Ted
Good. Now, substitute $\displaystyle u=x+1$.
Lets see what did you get.
• Jul 15th 2010, 10:09 AM
MechEng
$\displaystyle S=4\pi\frac{2(x+1)^{3/2}}{3}$$\displaystyle |_{1}^{4} \displaystyle S=\frac{8\pi(x+1)^{3/2}}{3}$$\displaystyle |_{1}^{4}$

$\displaystyle S=\frac{8\pi(4+1)^{3/2}}{3}-\frac{8\pi(1+1)^{3/2}}{3}$

$\displaystyle S=\frac{8\pi(5)^{3/2}}{3}-\frac{8\pi(2)^{3/2}}{3}$

$\displaystyle S=\frac{8\pi5\sqrt{5}}{3}-\frac{8\pi2\sqrt{2}}{3}$

$\displaystyle S=\frac{8\pi(5\sqrt{5}-2\sqrt{2})}{3}$

... Did I miss something?
• Jul 15th 2010, 10:20 AM
Ted
No. You are correct.