Take the 2-minute tour ×  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  $\displaystyle \frac{2}{3
}\int_0^{\pi/4}\left(\sec^2\theta+\frac{1}{4}\right)^{3/2}-
\frac{1}{8}d\theta$  $\displaystyle =\frac{2}{3}\int_0^{\pi/4}\l
eft(\sec^2\theta+\frac{1}{4}\right)^{3/2}d\theta-
\frac{\pi}{48}$.  As you can see, the integrand is stil