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Find the unit outward normal to each component of the region's boundary.

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By calculating a volume integral, find the volume $V$ of the region enclosed by the above surfaces.

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$V=$ [[0]].  (Enter your answer to 3d.p.)

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A region is enclosed by the surfaces $\\simplify{{a1}*x+{b1}*z}=\\var{c1}$, $x=\\var{d1}$, $y=\\var{e1}$, $y=\\var{f1}$, and $z=\\var{g1}$.

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Outward normals to the surfaces enclosing a region; volume of that enclosed region.

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a)

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The unit outward normals can most easily be identified by sketching the region bounded by the given surfaces which, in this case, is a wedge.

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Then the unit outward normals to the non-slanted surfaces are given by

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The outward normal to the final, slanted surface, is given by

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\\[\\boldsymbol{\\nabla}(\\simplify{{a1}*x+{b1}*z-{c1}})=\\pmatrix{\\var{a1},0,\\var{b1}},\\]

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and so the unit outward normal is given by

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\\[\\frac{1}{\\sqrt{(\\var{a1})^2+(\\var{b1})^2}}\\pmatrix{\\var{a1},0,\\var{b1}}=\\pmatrix{\\var{gradfhat[0]},\\var{gradfhat[1]},\\var{gradfhat[2]}}.\\]

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b)

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The volume of a region $V$ bounded by some particular surfaces is given by

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\\[V=\\int_V\\mathrm{d}x\\mathrm{d}y\\mathrm{d}z.\\]

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The relevant integral for the wedge in this question is therefore

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\\[V=\\int_{x=\\var{d1}}^{\\simplify{{c1-b1*g1}/{a1}}}\\mathrm{d}x\\int_{y=\\var{e1}}^{\\var{f1}}\\mathrm{d}y\\int_{z=\\var{g1}}^{\\simplify{{c1}/{b1}-{a1}/{b1}*x}}\\mathrm{d}z.\\]

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The integrals in $y$ and $z$ are straight forward, and we are left with

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\\[\\begin{align}V&=\\var{f1-e1}\\int_{x=\\var{d1}}^{\\simplify{{c1-b1*g1}/{a1}}}\\left(\\simplify{{-a1}/{b1}*x+{c1}/{b1}-{g1}}\\right)\\mathrm{d}x\\\\&=\\var{f1-e1}\\left[\\simplify{{-a1}/{2*b1}*x^2+{c1-g1*b1}/{b1}*x}\\right]_{x=\\var{d1}}^{\\simplify{{c1-b1*g1}/{a1}}}\\\\&=\\var{f1-e1}\\left\\{\\left(\\simplify{{-(c1-b1*g1)^2}/{2*a1*b1}+{(c1-g1*b1)^2}/{a1*b1}}\\right)-\\left(\\simplify{{-a1*d1^2}/{2*b1}+{d1*(c1-g1*b1)}/{b1}}\\right)\\right\\}\\\\&=\\var{vol}\\;\\text{to 3d.p.}\\end{align}\\]

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