# Compute the surface area of an oblate paraboloid

Consider the surface S: $z=4-4x^2-y^2, z\geq0$. Compute its surface area.

1. 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?

-

The intersection of the paraboloid with the xy-plane is $4x^2 + y^2 = 4$, so your polar variables substitution should be $x = r \cos\theta, y = 2r \sin\theta$. Your integral becomes $4 \int_0^{\frac{\pi}{2}} \int_0^{1} \sqrt{48 r^2 \cos^2 \theta + 17} r dr d\theta$, since we can exploit the symmetry of the surface. So saying, we have something which has no closed-form solution. You'll notice (by checking around the internet, or even here at M.SE) that the paraboloids in typical multivariate calculus problems have circular cross-sections. This lets the integrand simplify nicely -- or allows a Cavalieri-type approach using circular surface bands to work cleanly. Unfortunately, the circumference of an ellipse does not have a simple formula, so we get no help there (and your suggestion of using parabolic slices of surface area runs into difficulties also).