Discussion
The surface area of the portion of the paraboloid
\[ z = x^{2} + y^{2} \]
that lies between the planes \(z = 0\) and \(z = \frac{1}{4}\) is
Answer: A
The surface area is \(\frac{\pi}{6}(2\sqrt{2} - 1)\).
For a surface \(z = g(x,y)\), the area element is \(\sqrt{1 + g_{x}^{2} + g_{y}^{2}}\,dA = \sqrt{1 + 4r^{2}}\,dA\). The portion below \(z = \frac{1}{4}\) lies over the disc \(r \le \frac{1}{2}\). In polar coordinates:
\[ \begin{aligned} S &= \int_{0}^{2\pi}\!\int_{0}^{1/2} \sqrt{1 + 4r^{2}}\, r\,dr\,d\theta \\ &= 2\pi \left[ \tfrac{1}{12}(1 + 4r^{2})^{3/2} \right]_{0}^{1/2} \\ &= \tfrac{\pi}{6}\left(2^{3/2} - 1\right) = \tfrac{\pi}{6}(2\sqrt{2} - 1) \end{aligned} \]
The other options carry the same bracket with a wrong factor: \(\frac{\pi}{2}\), \(\frac{\pi}{3}\) and \(\pi\) come from losing the \(\frac{1}{12}\) of the substitution \(u = 1 + 4r^{2}\) or the \(\frac{1}{2}\) radius limit.
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