Find the area of the surface.
The part of the surface $z = 25 - x ^ { 2 } - y ^ { 2 }$ that lies above the $x y$-plane.
A) $( \sqrt { 626 } - 626 )$
B) $\frac { 2 } { 75 } ( 626 \sqrt { 626 } + 1 )$
C) $\frac { 2 } { 75 } \pi ( 626 \sqrt { 626 } - 1 )$
D) $\frac { 2 } { 75 } ( \sqrt { 626 } + 1 )$
E) $\pi ( 626 \sqrt { 626 } - 1 )$

Question

Quizplus · Accepted Answer

The surface is a downwards facing paraboloid, so we need to find the area of the part that lies above the $xy$-plane. We can use surface area integrals to find the area:
$$\iint_S dS = \iint_D \sqrt{1+\left(\frac{\partial z}{\partial x}ight)^2+\left(\frac{\partial z}{\partial y}ight)^2}dA$$
where $D$ is the projection of $S$ onto the $xy$-plane. 
We have $\frac{\partial z}{\partial x}=-2x$ and $\frac{\partial z}{\partial y}=-2y$, so
\begin{align*}
\iint_S dS &= \iint_D \sqrt{1+4x^2+4y^2}dA \
&= \int_0^{2\pi}\int_0^5 \sqrt{1+4r^2}rdrd	heta \
&= \frac{2}{75}\pi(626\sqrt{626}-1)
\end{align*}
Therefore, the answer is $\boxed{	extbf{(C)}}$.

Find the Area of the Surface z=25−x2−y2z = 25 - x ^ { 2 } - y ^ { 2 }z=25−x2−y2

Find the Area of the Surface $z = 25 - x ^ { 2 } - y ^ { 2 }$