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.
$$\int_{0}^{2\pi} \sqrt{\sin^4\theta+\sin^2\theta+2}d \,\theta$$
Integral $\int_{0}^{2\pi} \sqrt{\sin^4\theta+\sin^2\theta+2}\,d \theta$
definite integralselliptic integralsintegration
Best Answer
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 $2^{5/2}E(3/8)=7.98252\dots$