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\centerline
{
\bf Dr. Z's Math251 Handout \#15.6 [Surface Area]
}

By Doron Zeilberger

{\bf Problem Type 15.6a}:Find the area of the surface
$z=F(x,y)$ that lies above a given region.

{\bf Example Problem 15.6a}: 
Find the area of the surface
$z=1+3x+2y^2$ that lies above the triangle with vertices
$(0,0)$,$(0,1)$,$(2,1)$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Write the region either as a type I region
$$
D=\{ (x,y) \, \vert  \, a \leq x \leq  b \, ,f_1(x) \leq y \leq  f_2(x) \, \}
\,\, ,
$$
or type II region
$$
D=\{ (x,y) \, \vert  \, c \leq y \leq  d \, ,g_1(y) \leq x \leq  g_2(y) \, \}
\,\, ,
$$
whatever is convenient.
&
{\bf 1.} The triangle with vertices $(0,0)$,$(0,1)$,$(2,1)$
is most conveniently written as a type II region
$$
D=\{ (x,y) \, \vert  \, 0 \leq y \leq  1 \, ,0 \leq x \leq  2y \, \}
\quad .
$$
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{\bf 2.} Write the surface as $z=F(x,y)$ (if not already in that form)
and set-up the integral for the surface area
$$
\int\int_{D} \sqrt{1+ \left ( {{\partial z} \over {\partial x}} \right )^2
+ \left ( {{\partial z} \over {\partial y}} \right )^2 } 
\, dA \quad.
$$
Compute ${{\partial z} \over {\partial x}}=F_x(x,y)$
and ${{\partial z} \over {\partial y}}=F_y(x,y)$ and plug them in.
Then use the description of $D$ to convert it into an iterated integral.
&
{\bf 2.} In this problem, $F(x,y)=1+3x+2y^2$.
So $F_x=3$ and $F_y=4y$ and the integral is
$$
\int\int_{D} \sqrt{1+ 3^2+(4y)^2 } \, dA
$$
$$
=\int_{0}^{1}\int_{0}^{2y} \sqrt{10+16y^2 } \, dx \, dy
\quad .
$$
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{\bf 3.} Evaluate that iterated integral by first doing
the inner integral and then the outer integral.

&
{\bf 3.} The inner integral is
$$
\int_{0}^{2y} \sqrt{10+16y^2 } \, dx=
\sqrt{10+16y^2 } \int_{0}^{2y}  \, dx
$$
$$
=2y \sqrt{10+16y^2 } \quad ,
$$
and the outer integral is
$$
\int_{0}^{1}\int_{0}^{2y} \sqrt{10+16y^2 } \, dx \, dy
$$
$$
=\int_{0}^{1}
\left [ \int_{0}^{2y} \sqrt{10+16y^2 } \, dx \right ] \, dy
$$
$$
=\int_{0}^{1} 2y (10+16y^2)^{1/2} \, dy
= {{1} \over {16}} {{(10+16y^2)^{3/2}} \over {3/2}}
\Bigl \vert_{0}^{1}
$$
$$
={{1} \over {24}} \left [ (10+16 \cdot 1^2)^{3/2}-
(10+16 \cdot 0^2)^{3/2} \right ]
$$
$$
={{26^{3/2}-10^{3/2}} \over {24}} \quad.
$$
{\bf Ans.:} ${{26^{3/2}-10^{3/2}} \over {24}}$.

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