// Numbas version: exam_results_page_options {"name": "Find normals to surfaces enclosing a region, and the volume of the region", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "variables": {"b1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "b1"}, "theta2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "pi/2", "description": "", "name": "theta2"}, "a1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "name": "a1"}, "vol": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(0.5*b1*(theta2-theta1)*a1^2,3)", "description": "", "name": "vol"}, "theta1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "0", "description": "", "name": "theta1"}}, "ungrouped_variables": ["a1", "vol", "theta1", "b1", "theta2"], "name": "Find normals to surfaces enclosing a region, and the volume of the region", "functions": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

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

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

\n \n

 

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

\n

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

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A region is enclosed by the surfaces $x^2+y^2\\leqslant\\var{a1^2}$, $0\\leqslant z\\leqslant\\var{b1}$, $x\\geqslant 0$, $y\\geqslant 0$.

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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 quarter cylinder.

\n

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

\n \n

 

\n

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

\n

\\[\\boldsymbol{\\nabla}(\\simplify{x^2+y^2-{a1^2}})=\\pmatrix{2x,2y,0},\\]

\n

and so the unit outward normal is given by

\n

\\[\\frac{1}{\\sqrt{x^2+y^2}}\\pmatrix{x,y,0}.\\]

\n

 

\n

b)

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

\n

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

\n

which in this case is

\n

\\[V=\\int_R\\left[z\\right]_0^{\\var{b1}}\\mathrm{d}x\\mathrm{d}y,\\]

\n

where $R$ is the projection of $V$ onto the $\\pmatrix{x,y}$-plane.

\n

Due to the nature of the surface, however, it is convenient to simplify the integrals in $x$ and $y$, by using polar coordinates

\n

\\[\\begin{align}x&=r\\cos(\\theta),\\\\y&=r\\sin(\\theta),\\end{align}\\]

\n

then the integral becomes

\n

\\[V=\\var{b1}\\int_{\\var{theta1}}^{\\frac{\\pi}{2}}\\mathrm{d}\\theta\\int_0^{\\var{a1}}r\\mathrm{d}r,\\]

\n

which, after some straight forward integration, gives

\n

\\[V=\\simplify{{b1*a1^2}/4}\\pi=\\var{vol}\\;\\text{to 3d.p.}\\]

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