Find the mass of the surface $S$ having the given mass density.
$S$ is part of the plane $x + 5 y + 9 z = 45$ in the first octant; the density at a point $P$ on $S$ is equal to the square of the distance between $P$ and the $x y$-plane.
A) 42
B) $\sqrt { 107 }$
C) 19
D) $\frac { 375 \sqrt { 107 } } { 4 }$

Question

Quizplus · Accepted Answer

We can use a parameterization of the plane $S$ as $r(u,v) = \langle u, v, \frac{45- u - 5v}{9} \rangle$ for $0\leq u\leq 9$ and $0\leq v\leq \frac{45- u}{5}$. Then the mass of $S$ is given by the surface integral
$$\iint_S \rho \ dS = \iint_D \rho(r(u,v)) \|{\bf r_u}\times {\bf r_v}\| \ du \ dv$$
where $D$ is the parameter domain of $r(u,v)$. To find the density, we use the formula distance from a point to a plane, which gives
$$\text{density } = (9z)^2 = 81(45-x-5y)^2.$$
Substituting this and simplifying, we get 
$$\iint_S \rho \ dS = 81\iint_D (45-u-5v)^2 \sqrt{1^2+5^2+9^2} \ du \ dv.$$
We can evaluate this iterated integral in the usual way. First integrating with respect to $u$, we get
$$\int_0^9 (45-u-5v)^2 \ du = \frac{(23-v)^3}{3}\%$$
and then integrating with respect to $v$, we get
\begin{align*}
&\int_0^9 \frac{(23-v)^3}{3} \cdot 5\sqrt{107} \ dv\
&= 5\sqrt{107} \cdot \left( -\frac{(23-v)^4}{12} \right)\bigg|_0^9 \
&= \frac{375 \sqrt{107}}{4}.
\end{align*}
Therefore, the mass of $S$ is $\boxed{\textbf{(D) } \frac{375 \sqrt{107}}{4}}$.

Find the Mass of the Surface SSS Having the Given Mass Density

Find the Mass of the Surface $S$ Having the Given Mass Density