# Lateral surface area by hand

I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.

I have come up with my own solution, but I'm not sure if it's right.

Before substituting values back in, I got

$$\frac\pi{16} \left( (1+4x^2)^{3/2} 2x - \frac32 \ln\sqrt{1+4x^2+2x} + x\sqrt{1+4x^2} \right) \text{ from }0\text{ to }\sqrt2$$

Can someone please verify this? For my integration by parts, I set $u = \sec^3\theta - \sec\theta$ and $dv = \sec^2\theta d\theta$

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A generally useful check: if you have a tentative antiderivative, differentiate that and verify that the expression you have is identical to your original integrand. – J. M. Oct 4 '10 at 0:06

Not a complete answer, but maybe a bit of help:

The surface area is $\int_{x=0}^{\sqrt{2}} (2\pi y) ds = \int_{0}^{\sqrt{2}} 2 \pi x^2 \sqrt{1+(2x)^2} dx$, and the antiderivate is, according to Mathematica, $$\frac{\pi}{32} \left(2 \sqrt{4 x^2+1} \left(8 x^3+x\right)-\sinh^{-1}(2x)\right).$$ The inverse of $\sinh$ can be expressed using $\ln$, so you should be able to compare this to your expression.

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I assume you have in your earlier steps:

$\sigma = 2\pi \int_0^\sqrt{2} x^2 \sqrt{1+4x^2} dx$
$\tan \theta = 2x \; \mathrm{and} \; \sec \theta = \sqrt{1+4x^2}$
$\sigma = \frac{\pi}2 \int_{x=0}^{x=\sqrt{2}} \sec^3\theta \tan^2\theta d\theta.$

So you have a messy antiderivative you worked out, and want to check whether it's right? Why not see what happens if you take the derivative of your expression above? If everything is fine, you should get back to $2\pi x^2 \sqrt{1+4x^2}$. I get:

$d \sigma = \frac{\pi}{16} \left[ 24x^2\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - \frac{3(8x+2)}{4(1+4x^2+2x)} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx.$

Yes, there are a few problems here. More things should be cancelling out. The middle term (from $\ln$) is conspicuous for involving $\sqrt{1+4x^2+2x}$ instead of $\sqrt{1+4x^2}$. I'm pretty sure the $2x$ snuck under the radical by accident, and the term should really be $\ln | \sqrt{1+4x^2}+2x |$. If I fix that in your expression and derive again, I get

$d \sigma = \frac{\pi}{16} \left[ 24x\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - 3(1+4x^2)^{-\frac 12} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx.$

Now there are some like terms and related terms, but they won't cancel out far enough, because some of your coefficients are wrong. Maybe from here you can find your mistakes?

Other hints:

• The correct antiderivative in terms of $x$ never involves a factor of 3, unless you count the term $(1+4x^2)^{\frac 32}$.
• If you have a good antiderivative in terms of $\theta$, you might want to plug in to that equation instead of converting back to $x$, by finding values for $\tan \theta$ and $\sec \theta$ at $x=0$ and $x=\sqrt{2}$. (At $x=\sqrt{2}$, $\sec \theta$ is rational!)
• When you have a final answer, compare to my result $\frac{\pi}{32}\left[102\sqrt{2} - \ln\left(3+2\sqrt{2}\right)\right]$.
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With $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}\sec^2\theta d\theta$, $2\pi\int x^2\sqrt{1+4x^2}dx=\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta$. Now, let \begin{align} I_1&=\int\tan^2\theta\sec^3\theta d\theta \\ &=\int(\sec^3\theta-\sec\theta)\sec^2\theta d\theta \\ &=\tan\theta(\sec^3\theta-\sec\theta)-\int3\sec^3\theta\tan^2\theta-\sec\theta\tan^2\theta d\theta \\ &=\tan\theta(\sec^3\theta-\sec\theta)-3I_1+\int\sec\theta\tan^2\theta d\theta \end{align} (using the integration by parts as you suggested) so that $I_1=\frac{1}{4}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)$. Let \begin{align} I_2&=\int\sec\theta\tan^2\theta d\theta \\ &=\sec\theta\tan\theta-\int\sec^3\theta d\theta \\ &=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta d\theta-\int\sec\theta d\theta \\ &=\sec\theta\tan\theta-I_2+\log|\sec\theta-\tan\theta| \end{align} (using integration by parts with $u=\tan\theta$ and $dv=\sec\theta\tan\theta d\theta$) so that $I_2=\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)$. Now, \begin{align} \frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta &=\frac{\pi}{4}I_1=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right) \\ &=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)\right) \\ &=\frac{\pi}{16}\left(\tan\theta\sec^3\theta-\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\log|\sec\theta-\tan\theta|\right) \\ &=\frac{\pi}{16}\left(2x(1+4x^2)^{3/2}-x\sqrt{1+4x^2}+\frac{1}{2}\log\left|\sqrt{1+4x^2}-2x\right|\right) \end{align} which is similar to what you have, but not the same.

(Of course, doing this by hand and covering a sheet of paper with scribbles to do the two instances of over-and-back integration by parts, I may well have made a mistake along the way, but my result does appear to check by the method described in aschepler's answer.)

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