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  <question type="description">
    <name>
      <text>27.3.4.7 engineering example</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h4 class="HELM_example">Engineering Example</h4>
<h5> Radioactive  decay</h5>

<h6> Introduction</h6>

A cube of an impure radioactive ore is of side 10 cm. The number of radioactive decays taking place per cubic metre per second is given by \(R = 10^{23}(0.1-z)e^{-t/1000}\). The dependence on time represents a half-life of \(693\) seconds while the dependence on the vertical coordinate \(z\) represents some gravitational stratification. The value \(z=0\) represents the bottom of the cube and \(z=0.1\) represents the top of the cube. (Note that the dimensions are in metres so 10 cm becomes 0.1 m.) </p>
<p>What is the total number of decays taking place over the cube in the 100 seconds between \(t=0\) and \(t=100\)?</p> 


<h4 class="HELM_solution">Solution</h4>

The total number of decays is given by the quadruple
integral
\[\displaystyle 
N = \displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1} \displaystyle\int_{z=0}^{0.1}
\displaystyle\int_{t=0}^{100} 10^{23}(0.1-z)e^{-t/1000} \mathrm{d} t \,\mathrm{d}z \,\mathrm{d}y\,\mathrm{d}x\]  which may be evaluated as follows
\begin{align*}
N &= \displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1} \displaystyle\int_{z=0}^{0.1}
\displaystyle\int_{t=0}^{100} 10^{23}(0.1-z)e^{-t/1000} \mathrm{d} t \,\mathrm{d}z \,\mathrm{d}y\,\mathrm{d}x
\\
&= \displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1} \displaystyle\int_{z=0}^{0.1} \bigg[  -
1000 \times 10^{23}(0.1-z)e^{-t/1000} \bigg]_{t=0}^{100}  \;\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \\
 &= \displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1} \displaystyle\int_{z=0}^{0.1} \left[
  10^{26}(0.1-z)(1-e^{-0.1}) \right]  \;\mathrm{d}z \,\mathrm{d}y\,\mathrm{d}x \\
 &= \displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1} \displaystyle\int_{z=0}^{0.1} \left[
  9.5 \times 10^{24}(0.1-z) \right]  \;\mathrm{d}z \,\mathrm{d}y\,\mathrm{d}x \\
 &= 9.5 \times 10^{24}\displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1}  \bigg[
  (0.1z-0.5z^2) \bigg]_{z=0}^{0.1}  \;\mathrm{d}y\,\mathrm{d}x \\
 &= 9.5 \times 10^{24}\displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1}  \left[
  0.005 \right]  \;\mathrm{d}y\,\mathrm{d}x \\
 &= 0.005 \times 9.5 \times 10^{24}\displaystyle\int_{x=0}^{0.1} \displaystyle\int_{y=0}^{0.1}    \;\mathrm{d}y\,\mathrm{d}x \\
&= 0.005 \times 9.5 \times 10^{24} \times 0.1 \times 0.1 = 4.75
\times 10^{20}
\end{align*}
Thus the number of decays is approximately equal to \(4.75 \times
10^{20}\)]]></text>
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