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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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Enter your answer correct to 3 decimal places.

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Answer = [[0]]

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