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So im teaching an accelerated Calc II course. Which is asking to find the surface area of the surface created by rotating $y=\sin (\pi x)$ about the $x-$ axis from $0$ to $1$. So I know

$$S = 2 \pi \int_0^1 y \sqrt{1+(y')^2}dx.$$

So I end up with $y' = \pi \cos(\pi x)$ thus $(y')^2=\pi^2 \cos^2(\pi x)$.

But I'm stuck trying to solve

$$\int \sin (\pi x) \sqrt{1+\pi^2 \cos^2 (\pi x)}dx$$

IF I let $u=\cos x$ then the answer will be negative as $du$ will be negative. But I gave it a go and obtained

$$2 \int_{-1}^1 \sqrt{1+\pi^2u^2}du$$

Here I get stuck. So the answer is $2 \sqrt{1+\pi^2}$.

MyMathYourMath
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  • @GerryMyerson I haven’t taken calc since 08/09. I took some years off math then Returned to a PhD – MyMathYourMath Jan 19 '23 at 01:49
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    The substitution $x=a^{-1}\tan v$ leads to $\int\sec^3v,dv$. That's a hard one, but surely it has come up on this site before. – Gerry Myerson Jan 19 '23 at 01:50
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    For example, https://math.stackexchange.com/questions/154900/indefinite-integral-of-secant-cubed-int-sec3-x-dx – Gerry Myerson Jan 19 '23 at 01:52
  • If it's an accelerated Calculus II course most of the students will probably be better at integration than you are at present. So it's good that you are trying to fix this before the class (I hope it is before the class). Stick with it. – David Jan 19 '23 at 02:01
  • @David yes its before the class session lol – MyMathYourMath Jan 19 '23 at 02:33

2 Answers2

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"If I let $u=\cos(\pi x)$ then the answer will be negative" - no it won't. The integrand will be negative but the limits of integration will be reversed so the final answer will be positive. This is clear from the following line in your question. But if it bothers you, you could take $u=-\cos(\pi x)$ instead. Even better, $u=-\pi\cos(\pi x)$.

This will give an integral $$\int_{-a}^a \sqrt{1+u^2}\,du$$ times a constant - please find the constant yourself, also the value of $a$. Since the integrand is even, this is $$2\int_0^a \sqrt{1+u^2}\,du\ .$$ The best substitution for this is $u=\sinh\theta$. The answer should be $$I=\frac2\pi\,\sinh^{-1}\pi+2\sqrt{1+\pi^2}$$ which you can confirm on Wolfram Alpha if you wish.

David
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Let $u=\frac{1}{\pi}\tan \theta$, then $du=\sec^2 d\theta$. $$ \begin{aligned} \int_{-1}^1 \sqrt{1+\pi^2 u^2} d u & =2 \int_0^1 \sqrt{1+\pi^2 u^2} d u \\ & =2 \int_0^{\tan ^{-1} \pi} \sqrt{1+\tan ^2 \theta} \cdot \frac{1}{\pi} \sec ^2 \theta d \theta \\ & =\frac{2}{\pi} \int_0^{\tan ^{-1} \pi} \sec ^3 \theta d \theta \end{aligned} $$ Let’s deal with the $\sec^3\theta$ by integration by parts.

$$ \begin{aligned} \int\sec ^3 \theta d \theta & =\int \sec \theta d(\tan \theta) \\ & =\sec \theta \tan \theta-\int \sec \theta \tan ^2 \theta d \theta \\ & =\sec \theta \tan \theta-\int \sec \theta\left(\sec ^2 \theta-1\right) d \theta \\ \int \sec ^3 \theta d \theta & =\frac{1}{2}(\sec \theta \tan \theta+\ln |\sec \theta+\tan \theta|)+C \end{aligned} $$ Putting the limits back yields

$$ I=\frac{1}{\pi}[\sec \theta \tan \theta+\ln |\sec \theta+\tan \theta|]_0^{\tan ^{-1} \pi}=\sqrt{1+\pi^2}+\frac{1}{\pi} \ln \left| \pi+\sqrt{1+\pi^2} \right| $$

Lai
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