# Thread: Stumped on an integral.

1. ## Stumped on an integral.

Been beating my head against the wall on an integral for awhile now....

The question is this:

Compute the surface area generated by revolving the curve $\displaystyle $f(x) = 2\sqrt x ,\,\,{\text{for 1}} \leqslant x \leqslant 4$$ about the x-axis.

I took the 1st derivative of the equation:

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

Easy enough. Then I went into my integral:

$\displaystyle $2\pi \int_{\,1}^{\,4} {2\sqrt x \sqrt {1 + \frac{1}{{{x^2}}}} } dx$$

That didn't look very pretty, and I didn't seem to be getting anywhere trying to solve it (no u-sub, and integration by parts didn't seem to be working), so I tried to manipulate the equation several times to see if a different variation would work...

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

and

$\displaystyle $4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {\left( {{x^2} + 1} \right)} dx$$

and

$\displaystyle $4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} dx$$

and

$\displaystyle $4\pi \int_{\,1}^{\,4} {x\sqrt {\left( {\frac{{{x^2} + 1}}{{{x^3}}}} \right)} } dx$$

But I can't find one to work out. My professor stated on the assignment that the integral can be done exactly, otherwise I would have approximated it a long time ago.

Would one of you be able to possibly hint where I'm going wrong, or which form you'd take?

2. Here's a similar example for your consideration.

Areas of Surfaces of Revolution

3. Originally Posted by Malaclypse
Been beating my head against the wall on an integral for awhile now....

The question is this:

Compute the surface area generated by revolving the curve $\displaystyle $f(x) = 2\sqrt x ,\,\,{\text{for 1}} \leqslant x \leqslant 4$$ about the x-axis.

I took the 1st derivative of the equation:

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

Easy enough. Then I went into my integral:

$\displaystyle $2\pi \int_{\,1}^{\,4} {2\sqrt x \sqrt {1 + \frac{1}{{{x^2}}}} } dx$$

That didn't look very pretty, and I didn't seem to be getting anywhere trying to solve it (no u-sub, and integration by parts didn't seem to be working), so I tried to manipulate the equation several times to see if a different variation would work...

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

and

$\displaystyle $4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {\left( {{x^2} + 1} \right)} dx$$

and

$\displaystyle $4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} dx$$

and

$\displaystyle $4\pi \int_{\,1}^{\,4} {x\sqrt {\left( {\frac{{{x^2} + 1}}{{{x^3}}}} \right)} } dx$$

But I can't find one to work out. My professor stated on the assignment that the integral can be done exactly, otherwise I would have approximated it a long time ago.

Would one of you be able to possibly hint where I'm going wrong, or which form you'd take?
Your calculation of f'(x) is wrong.

4. The derivative of $\displaystyle ln(x)$ is $\displaystyle \frac{1}{x}$, not the derivative of $\displaystyle x^{1/2}$. To differentiate that, use the power rule.

5. Well now I feel like an idiot. Sure enough, I forgot to place the radical in the denominator when I performed it and didn't give it second thought. 1 / sqrt(x), NOT 1/x. I even looked at it several times. Urgh. Sorry all, and thanks.

That makes the integral nice and easy. What a waste of 3 hours.