Consider the surface S: $z=4-4x^2-y^2, z\geq0$. Compute its surface area.
- I've tried the following: $Area(S)=\int\int_D \sqrt{(8x)^2+(2y)^2+1}dxdy$ with D being the interior of the ellipse $x^2+(y/2)^2\leq 1$, and then switching to polar coordinates. At this point I had two strategies.
1.a Make the integrand simple: $x=r\cos\theta, y=4r\sin\theta$, and get Area(S)=$\int\int_{D_1}\sqrt{64r^2+1} 4rdrd\theta$ with D1: $\theta\in[0,2\pi]$, $0\leq r\leq \frac{1}{\sqrt{1+3\sin^2\theta}}$. This ends up as a bad integral in $\theta$: $\frac{1}{24}\left[4\int_0^{\pi/2}\left(\frac{64}{1+3\sin^2\theta}+1\right)^{3/2}d\theta-2\pi\right]$
1.b Make the domain nice: $x=r\cos\theta, y=2r\sin\theta$, but this turns out to be a horrible double integral (although, now over a rectangular domain)
2) I also tried a variation of the Cavalieri's principle: to compute Area(S), integrate the lengths of vertical sections (parts of parabolas). This ends up as a variation of integrals in 1.a and 1.b.
Any other suggestions for this problem? Or a way to evaluate the last integral in 1.a?