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<title>
Triple Integrals [11.7] 
and Volumes
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<h1>
Triple Integrals [11.7]
</h1>
<p class="show"><img src="0g/12.7.BlocksExplored.jpg" class="big">

<h3></h3>
<p class="show"><img src="0g/12.7.Blocks.jpg" style="width:320px" class="rightalign">
<img src="0g/12.7.Box.jpg"style="width:270px"><br>
Dividing a box, $B$...
<ul class="else">
<li> $l$ equal slices $\perp$ to $\uv i$,
<li> $m$ equal slices $\perp$ to $\uv j$,
<li> $n$ equal slices $\perp$ to $\uv k$,
</ul>


<br clear="both">
<p>The volume of one little block is
$$\Delta V = \Delta x \Delta y \Delta z.$$
<p><img src="0g/12.7.triple.jpg" class="big">
<p>And we can evaluate a triple integral by successive iterated integrals.<br>
<img src="0g/12.7.Fubini.triple.jpg" class="big"><br>
Order doesn't matter for a <b>box</b> integral, but one possibility is...
<ol type="incremental"><li>Evaluate...
$$A(y,z)=\int_a^b f(x,y,z)\,dx$$
while treating $y$ and $z$ as constants,
<li>$$B(z)=\int_c^d A(y,z)\,dy$$
while treating  $z$ as constant,
<li>$$\iiint_Bf(x,y,z)\,dV=\int_r^s B(z)\,dz.$$
</ol>


<h3>Integrating over a general bounded region, $E$</h3>
<p class="show">$u_1(x,y)$ and $u_2(x,y)$ are surfaces.<br>
$h_1(y)$ and $h_2(y)$ are curves / arcs in the $xy$ plane.<br><br>
<img src="0g/12.7.arbitraryV.jpg" class="rightalign">
Extend the idea of manipulating the limits of the integrals to set up triple integrals over some general bounded 3D region, $E$, whose boundaries can be described in functional form, e.g.
$$\iiint_Ef(x,y,z)\,dV=
\int_{y=c}^d\int_{x=h_1(y)}^{h_2(y)} \int_{z=u_1(x,y)}^{u_2(x,y)}f(x,y,z)\,dz\,dx\,dy.$$
Now the order of evaluating integrals (from inside to outside) does matter.

<p>To find the volume of the solid, $E$, defined by these limits:
$$V_E=
\int_{y=c}^d\int_{x=h_1(y)}^{h_2(y)} \int_{z=u_1(x,y)}^{u_2(x,y)}\,dz\,dx\,dy$$
where $dx\,dy\,dz=dV$.



<h3>To Do</h3>
<ul class="show todo">
<li>11.7 Volume problems
</ul>




<h4 class="handout">Image credits</h4>
<p class="handout"><a href="https://www.flickr.com/photos/realsmiley/4895250473">Mathias Rhomberg</a>


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