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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?

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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).

Here is what the general solution to your problem looks like -- you are crossing into elliptic-integral territory!

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