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 paraboloid with its vertex at (0,0,25). The radius of the base is $\sqrt{626}$, which can be found by setting $z=0$ and solving for $r=\sqrt{x^2+y^2}$. Thus, the surface area is given by $A=\iint_S dS=\iint_D ||r_u 	imes r_v|| dA$, where $r(u,v)=\langle u,v,25-u^2-v^2angle$ and $D$ is the disk in the $xy$-plane with radius $\sqrt{626}$ centered at the origin. We have
$$
r_u 	imes r_v = \begin{vmatrix} i & j & k\ 1 & 0 & -2u\ 0 & 1 & -2v \end{vmatrix} = \langle 2u,2v,1angle,
$$
so $||r_u 	imes r_v|| = \sqrt{4u^2+4v^2+1}$. Thus,
$$
A=\iint_S dS = \iint_D \sqrt{4u^2+4v^2+1} dA.
$$
Switching to polar coordinates, we have $x=r\cos	heta$ and $y=r\sin	heta$, so $dA=rdrd	heta$. The integral becomes
\begin{align*}
A &= \int_0^{2\pi}\int_0^{\sqrt{626}} \sqrt{4r^2+1} \cdot r drd	heta \
&= 2\pi \int_0^{\sqrt{626}} (2r^2+r\sqrt{4r^2+1}) dr \
&= 2\pi \left[ \frac{1}{3}(2r^3+3r\sqrt{4r^2+1})ight]_0^{\sqrt{626}} \
&= \frac{2}{75}\pi(626\sqrt{626}-1).
\end{align*}

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 $z = 25 - x ^ { 2 } - y ^ { 2 }$