# Surface area of revolution problems without perfect squares

When calculating surface area of revolution, I often find myself in situations like this :

Problem:

My work:

$f(x)=\frac{x^3}{5}$

$f'(x) = \frac{3x^2}{5}$

$f'(x)^2 = \frac{9x^4}{25}$

$1 + f'(x) = \frac{25}{25} + \frac{9x^4}{25} = \frac{9x^4 + 25}{25}$

SA = $2\pi \int_0^2 \frac{x^3}{5}\sqrt{\frac{9x^4+25}{25}} = \frac{\sqrt{9x^4+25}}{5}$

And this is where I stall out: In some problems, you end up with perfect squares in the numerator and denominator and the equation becomes easy to integrate. What are some strategies to employ when this is not the case - such as in this example? I have a feeling that completing the square is the trick here, but I haven't used it since pre-calc and I can't quite wrap my head around how to use it here.

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Hint: We need to find $$\int_0^2 \frac{2\pi x^3}{5} \sqrt{25+9x^4}\,dx.$$ Make the substitution $25+9x^4=u$.

We get $du =36x^3\,dx$, so $x^3\,dx=\frac{1}{36}\,du$. The rest should be straightforward. We end up needing to integrate a constant times $u^{1/2}$

Remark: If you take a "random" function, and set up the integral for arclength or surface area, you will usually end up with a function that does not have an elementary antiderivative. That is why the functions in arclength/surface area problems are usually very carefully chosen.

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In fact, out here in the real world (as opposed to calculus-homework-world) functions hardly ever have nice anti-derivatives. – bubba Mar 6 '14 at 3:42
Omitting the $x$ term was a typo, but the $2\pi$ is there, just moved outside the integral. Please see my edit and confirm that I have the integral set up correctly. Also, my weakness here is in the u-substitution - it's a subject I didn't fully grasp in calc 1: I'm trying to grasp it now, and it'd help a lot if you could tell me what u and du should be in this specific instance. – aqaeous Mar 6 '14 at 3:44
Thanks, that helps a lot. Just to make sure I have it straight, the 5 in the denominator under $2\pi x^3$ was originally from the given $y=\frac{x^3}{5}$, and there was a 25 in the denominator under the radical. We took the root of that denominator to get this: $\int_0^2 \frac{2\pi x^3}{5}\frac{\sqrt{25+9x^4}}{5}$. We pull that 5 out by multiplying the equation by $\frac{1}{5}$, which then "consolidates" into the leading term? – aqaeous Mar 6 '14 at 4:46
I don't understand. The (displayed) integral for the surface area is the one in your current post. Note that I wrote we need to integrate a constant times $u^{1/2}$. The constant is $\frac{2\pi}{5\cdot 36}$, that's where the $5$ is. – André Nicolas Mar 6 '14 at 4:46
When you start out "raw" the square root term is $\sqrt{1+\frac{9x^4}{25}}$. We really don't need to simplify, we could let $u=1+\frac{9x^4}{25}$. But I went along with your (correct) simplification, which is to $\sqrt{\frac{25+9x^4}{25}}$, and then, pulling out the $\sqrt{25}$ as $5$, to the expression we end up with. – André Nicolas Mar 6 '14 at 4:50

Hint : $$S = \int_0^2 2\pi y\sqrt{[1+(\frac{dy}{dx})^{2}]}dx$$

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