Question

The surface area of the portion of the paraboloid \[ z=x^2+y^2 \] that lies between the planes $z=0$ and $z=\tfrac{1}{4}$ is

• A. $\dfrac{\pi}{6}\big(2\sqrt{2}-1\big)$
• B. $\dfrac{\pi}{2}\big(2\sqrt{2}-1\big)$
• C. $\pi\big(2\sqrt{2}-1\big)$
• D. $\dfrac{\pi}{3}\big(2\sqrt{2}-1\big)$

Hint:

For rotationally symmetric graphs $z=f(r)$, switch to polar coordinates early. The surface element becomes $\sqrt{1+(f'(r))^2}\,r\,dr\,d\theta$ with $f'(r)=2r$ here.

Answer 1

The Correct Option is A

Step 1: Set up the surface-area integral.
For $z=f(x,y)$, surface area over domain $D$ is \[ S=\iint_D \sqrt{1+f_x^2+f_y^2}\,dx\,dy. \] Here $f_x=2x,\ f_y=2y \Rightarrow \sqrt{1+f_x^2+f_y^2}=\sqrt{1+4(x^2+y^2)}$.
Step 2: Determine the projection domain.
$0\le z=x^2+y^2\le \tfrac14 \Rightarrow x^2+y^2\le \tfrac14$. In polar coordinates $(r,\theta)$: $0\le r\le \tfrac12,\ 0\le \theta\le 2\pi$.
Step 3: Evaluate in polar coordinates.
\[ S=\int_0^{2\pi}\!\!\int_0^{1/2} \sqrt{1+4r^2}\, r\,dr\,d\theta =2\pi\int_0^{1/2} r\sqrt{1+4r^2}\,dr. \] Let $u=1+4r^2 \Rightarrow du=8r\,dr$: \[ S=2\pi . \frac{1}{8}\int_{u=1}^{2} u^{1/2}\,du =\frac{\pi}{4}\left[\frac{2}{3}u^{3/2}\right]_{1}^{2} =\frac{\pi}{6}\big(2^{3/2}-1\big) =\frac{\pi}{6}\big(2\sqrt{2}-1\big). \] Final Answer:\fbox{$\dfrac{\pi}{6}\left(2\sqrt{2}-1\right)$}