I was calculating the area of this surface $\phi:(r \cos \theta, 2r \sin \theta, r)$and I was left with this integral but I have no idea how to solve it, I tried looking for it in Wolfram but I would like to know how to solve it by hand. Thanks for any hint or solution. $$\int_{0}^{2\pi} \sqrt{\sin^4\theta+\sin^2\theta+2}d \,\theta$$
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Not really related to the question, but what was the range of $r$ in the surface. – Thomas Andrews May 27 '21 at 23:26
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Not sure how helpful this is, but the integral is equivalent to $(1/2)\int_{0}^{2\pi}\sqrt{\sin^2\theta+4\sin\theta +11}d\theta$, which means its value is pretty close to $\pi\sqrt{23/2}$ – eyeballfrog May 27 '21 at 23:34
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Also, what does this surface look like? – Тyma Gaidash May 27 '21 at 23:59
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Did you notice that WA is doing it numerically? That typically doesn't bode well for an analytical solution. – Ian May 28 '21 at 00:20
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@Ian. There is an explicit antiderivative even for the original integrand (a monster) – Claude Leibovici May 28 '21 at 04:10
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@ClaudeLeibovici It's not that bad - I worked it out before. See my answer. – Parcly Taxel May 28 '21 at 07:31
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As mentioned in comments, the integral in question is equivalent to $$\frac12\int_0^{2\pi}\sqrt{\sin^2\theta+4\sin\theta+11}\,d\theta$$ By substituting $t=\sin\theta$ we turn this into $$\int_{-1}^1\sqrt{\frac{t^2+4t+11}{1-t^2}}\,dt$$ which is an instance of an integral appearing in this answer of mine with $c=\sqrt{-7}-2$. Substituting this in and simplifying gives the result in terms of the three elliptic integrals as $$2^{11/4}(E(m)-K(m)+(1-n)\Pi(n,m))\qquad m=\frac{8-5\sqrt2}{16},n=\frac{4-3\sqrt2}8$$ where arguments are as in Mathematica/mpmath.
However, this does not seem to be the surface area of the surface given in the question for $\theta\in[0,2\pi]$ and a nice value of $r$. That area, of an elliptical cone, for $r\in[0,1]$ is given by $2^{5/2}E(3/8)=7.98252\dots$
Parcly Taxel
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This is nice ! Thanks for posting it and $\to +1$ for sure. Cheers :-) – Claude Leibovici May 28 '21 at 07:36