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    <name>
      <text>29.2.4.17 div Vector field volume integral more advanced</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<p>For the vector field \(\underline{F} = (x^2y+\sin z)\underline{i}+({@aa*x*y^2@}+e^z )\underline{j}+(z^2+{@bb*x^y@})\underline{k}\), find the integral \(\displaystyle \iiint_V \underline{\nabla}\cdot \underline{F} \mathrm{d}V\) where \(V\) is the volume inside the tetrahedron bounded by \(x = 0\), \(y = 0\), \(z = 0\) and \(x+y+z = 1\).</p>

<p>\(\displaystyle \iiint_V \underline{\nabla}\cdot \underline{F} \mathrm{d}V =\) [[input:ans1]][[validation:ans1]].</p>]]></text>
    </questiontext>
    <generalfeedback format="moodle_auto_format">
      <text><![CDATA[<p>For the vector field \(\underline{F} = (x^2y+\sin z)\underline{i}+({@aa*x*y^2@}+e^z )\underline{j}+(z^2+{@bb*x^y@})\underline{k}\), we aim to find the integral \(\displaystyle \iiint_V \underline{\nabla}\cdot \underline{F} \mathrm{d}V\) where \(V\) is the volume inside the tetrahedron bounded by \(x = 0\), \(y = 0\), \(z = 0\) and \(x+y+z = 1\).</p>

<p>We will first calculate \(\underline{\nabla} \cdot \underline{F}\):
\[
\begin{aligned}
\underline{\nabla} \cdot \underline{F} &= \frac{\partial}{\partial x}(x^2y+\sin z) + \frac{\partial}{\partial y}({@aa*x*y^2@}+e^z ) + \frac{\partial}{\partial z}(z^2+{@bb*x^y@})\\
&= 2xy + {@2*aa*x*y@}+ 2z\\
&={@(2*aa+2)*x*y@}+2z
\end{aligned}
\]
Now Calculating the integral with the bounds defined in the question:
\[
\begin{aligned}
\iiint_V \underline{\nabla} \cdot \underline{F} \mathrm{d}V &= \int^{z=1}_{z=0} \int^{y=1-z}_{y=0} \int^{x=1-y-z}_{x=0} {@(2*aa+2)*x*y@}+2z\ \mathrm{d}x\ \mathrm{d}y\ \mathrm{d}z\\
&= \int^{z=1}_{z=0} \int^{y=1-z}_{y=0} \left[{@First_int_indef@}\right]^{x=1-y-z}_{x=0}\ \mathrm{d}y\ \mathrm{d}z\\
&= \int^{z=1}_{z=0} \int^{y=1-z}_{y=0}{@First_int_def@} \ \mathrm{d}y\ \mathrm{d}z\\
&= \int^{z=1}_{z=0} \left[{@Second_int_indef@}\right]^{y=1-z}_{y=0}\ \mathrm{d}z\\
&= \int^{z=1}_{z=0} {@Second_int_def@}\ \mathrm{d}z\\\\
&=\left[{@Third_int_indef@}\right]^{z=1}_{z=0}\\
&={@Third_int_def@}
\end{aligned}
\]</p>]]></text>
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      <text>2025073100</text>
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    <questionvariables>
      <text>aa:rand(5)+1;
bb:rand(5)+1;

expression:(2*aa+2)*x*y+2*z;
First_int_indef:integrate(expression, x);
First_int_def:integrate(expression, x, 0, 1-y-z);
Second_int_indef:integrate(First_int_def, y);
Second_int_def:integrate(First_int_def, y, 0, 1-z);
Third_int_indef:integrate(Second_int_def, z);
Third_int_def:integrate(Second_int_def, z, 0, 1);

ta:Third_int_def;</text>
    </questionvariables>
    <specificfeedback format="html">
      <text>[[feedback:prt1]]</text>
    </specificfeedback>
    <questionnote format="html">
      <text>\(\underline{F} = (x^2y+\sin z)\underline{i}+({@aa*x*y^2@}+e^z )\underline{j}+(z^2+{@bb*x^y@})\underline{k} \text{ so } \displaystyle \iiint_V \underline{\nabla}\cdot \underline{F} \mathrm{d}V = {@ta@}\)</text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:orange;"><i class="fa fa-adjust"></i></span> Your answer is partially correct.]]></text>
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