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<title>
Surface area
[11.6+] 
</title>
<meta name="week" content="10">
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<h1>
Surface Area [11.6]
</h1>

<p class="show">Coming on the heels of  <a href="mv11.6.php">Parametric Surfaces</a>...

<ul class="else">
<li>As parameters $u$ and $v$ are varied in some domain $D$,
<li>The vector
$\myv r(u,v)=\langle\,x(u,v),\,y(u,y),\,z(u,v)\rangle$
traces out a surface $S$ in 3D.


<h3>Tangent plane approximation</h3>
<p>Outline of the tangent plane approximation:<br>
<img src="0g/12.5.surfuv.jpg" class="big">
<ul><li>As $v$ varies in $D$, ($u$ is constant), $\myv r$ traces the curve $C_1$.  A tangent vector to $C_1$ is
$$\myv r_v
=\langle \frac{\del x}{\del v},
\frac{\del y}{\del v},\frac{\del z}{\del v}
\rangle\equiv \langle x_v, y_v, z_v \rangle.$$
[What does that subscript mean in this context??]
<li>As $u$ varies in $D$, ($v$ is constant), $\myv r$ traces the curve $C_2$.  A tangent vector to $C_2$ is
$$\myv r_u=\langle x_u, y_u, z_u \rangle$$

<li>Increasing $v$ to $v+\Delta v$ in $D$ leads to a change of $\myv r$ on the surface of <i>approximately</i> $\Delta v\, \myv r_v$.  
<li>Increasing $u$ to $u+\Delta u$ in $D$ leads to a change of $\myv r$ on the surface of <i>approximately</i> $\Delta u\, \myv r_u$.

<li><img src="0g/12.5.patch.jpg" class="rightalign">The area of the parallelogram shown is <br><br>
$|(\Delta v\,\myv r_v)\times(\Delta u\,\myv r_u)|$<br>
 &nbsp; =$|\myv r_v\times\myv r_u|\Delta A$<br>
<br>Where $\Delta u \Delta v=\Delta A$, an area in <b>nice, well behaved, orthonormal</b> $u \times v$ space.

<li>This cross product is <i>approximately</i> equal to $\Delta S$, the area of a "patch" of the surface. 

</ul>
<p>In the limit $\Delta u\to du,\ \Delta v \to du$, the approximation gets more exact, and..
<p class="def show incremental">The surface area  $S$ is
$$S=\iint_D dS = \iint_D |\myv r_u\times \myv r_v|\,dA=\iint_D |\myv r_u\times \myv r_v|\,du\,dv.$$

<p class="todo">
Do <a href="https://activecalculus.org/multi/S-11-6-Parametric-Surfaces-Surface-Area.html#A_11_6_4">A 11.6.3 - Surface area of a cylinder</a>.

<p id='surfacenormal'>[Later on, we'll consider <a href="13.6.php">surface integrals [13.6]</a>.  It will be useful that the direction of the cross-product $\myv r_u \times \myv r_v$ is perpendicular to (normal to) the surface.]


<h3>Surface area of a graph defined in terms of $(x,y)$</h3>
<p class="show">The general expression for surface area:
$$S=\iint_D dS = \iint_D |\myv r_u\times \myv r_v|\,dA.$$

<p>For the special case of the surface $z=f(x,y)$, $u\to x$, $v\to y$, such that $(x,y)$ is in the domain $D$,
$$\begineq \myv r_u\to \myv r_x &=\langle \frac{\del x}{\del x},
\frac{\del y}{\del x},\frac{\del z}{\del x}
\rangle\\
&=\langle 1,
0,\frac{\del z}{\del x}
\rangle\\
\endeq$$
Since $x$ and $y$ are at right angles (orthogonal), a change in $x$ does not change $y$, and vice versa.  So $\frac{\del x}{\del y}=
\frac{\del y}{\del x}=0$.<br><br>
And,

$$\begineq \myv r_v\to \myv r_y &=\langle 0,
1,\frac{\del z}{\del y}
\rangle\\

\endeq$$

<p>Writing out the cross product of the two vectors:
$$\begineq \myv r_x \times \myv r_y &=

\text{det}\begincv \hat i&\hat j& \hat k\\
1&0&\frac{\del z}{\del x }\\
0&1&\frac{\del z}{\del y }
\endcv
\\
&=
-\frac{\del z}{\del x }\hat i-\frac{\del z}{\del y }\hat j +\hat k.
\endeq$$

<p>And therefore the 
surface area can be written as:
$$A_S=\iint_D|\myv r_x\times \myv r_y|\,dA
=
\iint_D
\sqrt{1+\left(\frac{\del z}{\del x}\right)^2 + \left(\frac{\del z}{\del y}\right)^2
} \, dA
$$

<h4>Surface area of a graph defined in terms of $(x,y)$</h3>
<p>For the special case of the surface $z=f(x,y)$, $u\to x$, $v\to y$, such that $(x,y)$ is in the domain $D$, 
the surface area can be written (see *below)
:
$$S=\iint_D\sqrt{1+\left(\frac{\del z}{\del x}\right)^2 + \left(\frac{\del z}{\del y}\right)^2
} \, dA
$$

<h3 id="example">Example</h3>
<p class="show todo"><img src="0g/12.5.1-x-y.jpg" class="rightalign">
Calculate the area of the surface shown of the plane:<br>
$z=1-x-y$
<br>
<a href="#example" onclick="toggle_visibility('aexample');">show / hide</a>
<div id="aexample" class="result toggle">

<p>$$\begineq
S&=\iint_D\sqrt{1+\left(\frac{\del z}{\del x}\right)^2 + \left(\frac{\del z}{\del y}\right)^2
} \, dA
\\
&=\iint_D\sqrt{1+\left(-1\right)^2 + \left(-1\right)^2
} \, dA
=\iint_D\sqrt{3} \, dA
\\
&=\sqrt 3 \iint_D dA=\sqrt 3 A_D=\sqrt 3 *(1*1)/2=\sqrt 3/2
\endeq
$$
</div>

<p>Check your  result by using geometry to calculate the surface area of the colored triangle.



