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So I was digging in my old notes and found this teaser from a few months ago that neither the class nor the teacher could solve:

Find the surface area of the part of $z=4-x^2-y^2$ that lies above the triangle formed by $(0,0)$, $(1,0)$, and $(1,1)$.

Since the integral in rectangular (Cartesian) coordinates isn't particularly nice, I used cylindrical coordinates instead. Since it's simple enough to see that radially, the region goes from the origin out to $x=1=r\cos\theta$, and the line from the origin passing through $(1,1)$ forms an angle of $\theta=\pi/4$, the integral would just be $\displaystyle \iint_R\sqrt{1+z_x^2+z_y^2} dA = \int_0^{\pi/4}\int_0^{\sec\theta}r\sqrt{1+4r^2}drd\theta$.

With the power of math, this simplifies to

$\displaystyle \frac{2}{3}\int_0^{\pi/4}\left(\sec^2\theta+\frac{1}{4}\right)^{3/2}-\frac{1}{8}d\theta$

$\displaystyle =\frac{2}{3}\int_0^{\pi/4}\left(\sec^2\theta+\frac{1}{4}\right)^{3/2}d\theta-\frac{\pi}{48}$.

As you can see, the integrand is still messy. Am I going in the right direction, and if I am, how do you integrate this monster?

P.S. I'm marking this as homework because I can get what is presumably the correct answer, $\frac{1}{12}\left(6+\tan^{-1}(1/3)+7\tanh^{-1}(2/3)-\pi/4\right)\approx0.9307820903765$, from Mathematica. I'm just curious about how to get there.

[[2013/04/03 EDIT: the secant function can be replaced with its exponential form $1/\cosh(i\theta)$, so we have $\displaystyle =\frac{2}{3}\int_0^{\pi/4}\left(\frac{1}{4}+\frac{4\exp(2i\theta)}{(1+\exp(2i\theta))^2}\right)^{3/2}d\theta$. Still a bit lost though, from there.]]

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I'm afraid this is not a "nice" integration at all, regardless of the choice of coordinate system. As I recall, the hemispherical surface $ \ z \ = \ \sqrt{4-x^2-y^2} \ $ over this region is far more tractable; but the paraboloid is pretty awful. As one learns even with arclength problems, just because a curve (or, here, a surface) is fairly simple doesn't mean its length (or surface) integrals are. (Mathematica may have given you a definite integral with not too many terms, but the indefinite integral for this is... sizeable.) – RecklessReckoner Apr 3 '14 at 22:21

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