I've been dealing with a problem of surface integrals, I hope you could give me an advice.
I have this parameterized function $G = (r\cos\theta, r\cos\theta, r)$ with $0\le r \le 1$ and $0\le\theta\le2\pi$. I have to find a formula for the surface area of this parameterized function.
I started derivating $G$ so $$G_r = (\cos\theta, 2\sin\theta,1)$$ $$G_\theta = (-r\sin\theta, 2r\cos\theta, 0)$$ $$||G_r\times G_\theta|| = \sqrt{4r^2\cos^2\theta+r^2\sin^2\theta+4r^2} = \sqrt{r^2(3\cos^2\theta+5)}=r\sqrt{3\cos^2\theta + 5}$$ Now I tried to do the integral with the given integral limits of $r$ and $\theta$
$$\int_0^{2\pi}\int_0^{1}r\sqrt{3\cos^2\theta+5}drd\theta=\int_0^{2\pi}\cfrac{\sqrt{3\cos^2\theta+5}}{2}d\theta$$
I saw that the solution had to do with a complete elliptic integral, but I would hope you could guide me if there is another way to do that integral and also if I'm doing any procedure wrong.