Math Help - Integrate sqrt(1+(1\sqrt(x)))^2

1. 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:
$f(x) = 2\sqrt{x}$ for $1\leq x \leq4$

So far I have the following:

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

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

So...

$
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.

2. The problems wants you to find the surface area not the length of the curve.

3. Yes, it does want me to find the surface area. However, I still need to be able to integrate $\int_{1}^{4}{\sqrt{\frac{x+1}{x}} dx$ as $S=\int_{1}^{4}{2\pi\sqrt{\frac{x+1}{x}} dx$ is the same as $S=2\pi\int_{1}^{4}{\sqrt{\frac{x+1}{x}} dx$ ... No?

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

4. Originally Posted by MechEng
Is finding the surface not as simple as multiplying the lentgh of the curve by $2\pi$?
Becareful, you are multiplying the dS (which is $\sqrt{1+f^2(x)}$) not the length of the curve!

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

So the required surface area is:

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

Can you solve it?

5. I think so...

Jumping to the end, I get:

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

Does that look correct to you?

6. That 8 should be 4.

7. $S=\int_{1}^{4}{2\pi2\sqrt{x}\sqrt{\frac{x+1}{x}}} dx$

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

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

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

8. Good. Now, substitute $u=x+1$.
Lets see what did you get.

9. $S=4\pi\frac{2(x+1)^{3/2}}{3}$ $|_{1}^{4}$

$S=\frac{8\pi(x+1)^{3/2}}{3}$ $|_{1}^{4}$

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

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

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

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

... Did I miss something?

10. No. You are correct.