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\$$\\int\\int_R\\frac{\\var{a}}{\\var{t}\\sqrt{x^2+y^2}}dxdy\$$

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Limits

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\$$x^2+y^2\\ge\\simplify{{k}^2}\$$ and \$$x^2+y^2\\le\\simplify{{k2}^2}\$$.

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This is the area enclosed between two circles the first having radius \$$\\var{k}\$$ and the second with radius \$$\\var{k2}\$$ and both having centre (0, 0)

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\$$x=rcos(\\theta)\$$      and      \$$y=rsin(\\theta)\$$

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\$$\\var{k}\\le r\\le\\var{k2}\$$       and        \$$0\\le \\theta\\le2\\pi\$$

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\$$dxdy=rdrd\\theta\$$

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The function

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\$$x^2+y^2=r^2\$$

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\$$\\implies \\frac{\\var{a}}{\\var{t}\\sqrt{x^2+y^2}}dxdy=\\frac{\\var{a}}{\\var{t}\\sqrt{r^2}}=\\frac{\\var{a}}{\\var{t}r}\$$

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The Integral

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\$$\\int\\int_R\\frac{\\var{a}}{\\var{t}\\sqrt{x^2+y^2}}dxdy=\\int_0^{2\\pi}\\int_\\var{k}^{\\var{k2}}\\frac{\\var{a}}{\\var{t}r}rdrd\\theta=\\int_0^{2\\pi}\\int_\\var{k}^{\\var{k2}}\\frac{\\var{a}}{\\var{t}}drd\\theta\$$

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Inner integral

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\$$\\int_\\var{k}^{\\var{k2}}\\frac{\\var{a}}{\\var{t}}dr=\\frac{\\var{a}}{\\var{t}}r\\big|_\\var{k}^{\\var{k2}}\$$

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\$$=\\frac{\\var{a}}{\\var{t}}(\\var{k2})-\\frac{\\var{a}}{\\var{t}}(\\var{k})\$$

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\$$=\\frac{\\simplify{{c}*{a}}}{\\var{t}}\$$

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Outer Integral

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\$$\\int_0^{2\\pi}\\frac{\\simplify{{c}*{a}}}{\\var{t}}d\\theta\$$

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\$$=\\frac{\\simplify{{c}*{a}}}{\\var{t}}\\theta\\big|_0^{2\\pi}\$$

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\$$=\\frac{\\simplify{{c}*{a}}}{\\var{t}}({2\\pi})-0\$$

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\$$=\\simplify{{c}*{a}*2*{pi}/{t}}\$$

", "statement": "

Evaluate the integral below using polar co-ordinates:

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\$$\\int\\int_R\\frac{\\var{a}}{\\var{t}\\sqrt{x^2+y^2}}dxdy\$$

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where \$$R\$$ is the region of the plane enclosed by the circles \$$x^2+y^2\\ge\\simplify{{k}^2}\$$ and \$$x^2+y^2\\le\\simplify{{k2}^2}\$$.

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