// Numbas version: exam_results_page_options {"name": "Ida Friestad's copy of Use Green's theorem to convert line integral to double integral, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "g", "f"], "name": "Ida Friestad's copy of Use Green's theorem to convert line integral to double integral, ", "tags": ["Calculus", "calculus", "checked2015", "closed path", "differentiation", "Green's theorem", "green's theorem", "Greens theorem", "greens theorem", "integral over a closed path", "integral over a rectangle", "integral over a region", "line integral", "MAS2104", "partial differentiation", "tested1"], "type": "question", "advice": "

First we identify the functions $u$ and $v$ and their required derivatives:

\\[\\begin{eqnarray*} u &=& \\simplify[std]{{a}x^2-{b}y} \\Rightarrow \\frac{\\partial u}{\\partial y} = \\simplify[std]{-{b}} \\\\ v&=&\\simplify[std]{{c}y^2+{d}x} \\Rightarrow \\frac{\\partial v}{\\partial x} = \\simplify[std]{{d}}\\end{eqnarray*}\\]

Hence: \\[\\frac{\\partial v}{\\partial x}-\\frac{\\partial u}{\\partial y}=\\simplify[std]{{d}-{-b}}=\\var{d+b}.\\]

So, using Green’s Theorem, the integral $I$ becomes:

\\[\\begin{eqnarray*} I&=&\\int \\int_R\\left(\\frac{\\partial v}{\\partial x}-\\frac{\\partial u}{\\partial y}\\right)\\;dx\\;dy\\\\ &=&\\int \\int_R \\var{d+b}\\;dx\\;dy = \\var{d+b}\\int \\int_R dx\\;dy\\\\ &=&\\var{d+b}\\times \\textrm{Area of }R. \\end{eqnarray*}\\]

Now the region $R$ is a rectangle of size $\\var{f}$ by $\\var{g}$. Hence:

\\[\\begin{eqnarray*} I&=&\\var{b+d}\\times \\var{f} \\times \\var{g}\\\\ &=& \\var{(b+d)*f*g}. \\end{eqnarray*} \\]

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$I=\\;\\;$[[0]]

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Green’s theorem states that for a region R with boundary $\\Gamma$

\\[\\oint_{\\Gamma} \\left( u\\;dx+v\\;dy \\right)= \\int \\int_R\\left(\\frac{\\partial v}{\\partial x}-\\frac{\\partial u}{\\partial y}\\right)\\;dx\\;dy.\\] Use Green’s theorem to find the value of: \\[I=\\oint_{\\Gamma} \\left( \\left(\\simplify[std]{{a}x^2-{b}y} \\right)\\;dx+\\left(\\simplify[std]{{c}y^2+{d}x}\\right)\\;dy\\right)\\]

where $\\Gamma$, mapped counter-clockwise, is the closed path, starting at $(0,0)$, around the boundary of a rectangle with vertices $(0,0),\\;(\\var{f},0),\\;(\\var{f},\\var{g}),\\;(0,\\var{g})$.

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30/06/2012:

Added tags. Could include Show steps on Green's theorem.

19/07/2012:

Added Show steps on Green's Theorem.

Added description.

Checked calculation.

23/07/2012:

Added tags.

In the question and Steps added brackets so that Green's Theorem is valid.

Question appears to be working correctly.

23/12/2012: (WHF)

No Shoe steps on Green's theorem.

Checked calculations, OK. Added tested1 tag. Few minor typos - full stops added.

", "description": "

(Green’s theorem). $\\Gamma$ a rectangle, find: $\\displaystyle \\oint_{\\Gamma} \\left(ax^2-by \\right)\\;dx+\\left(cy^2+px\\right)\\;dy$.

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