Answer to Question #277190 in Calculus for Air

Question #277190

∫ ∫ π‘₯𝑦𝑑π‘₯𝑑𝑦 2𝑦


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1
Expert's answer
2021-12-09T16:03:13-0500

In unit circle we need to find the integral value.

"\\int \\int xydxdy= \\int_0^{2\\pi}\\int_0^1r^3sin\\theta cos\\theta drd\\theta\\\\\n=\\int_0^{2\\pi}[\\frac{r^4}{4}]_0^1sin\\theta cos\\theta drd\\theta\\\\\n=\\int_0^{2\\pi}\\frac{1}{4}sin\\theta cos\\theta drd\\theta\\\\\n=\\int_0^{2\\pi}\\frac{1}{8}2sin\\theta cos\\theta drd\\theta\\\\\n=\\int_0^{2\\pi}\\frac{1}{8}sin(2\\theta)drd\\theta\\\\\n=\\frac{1}{8}[\\frac{-cos(2\\theta)}{2}]_0^{2\\pi}\\\\\n=\\frac{1}{16}[-1+1]\\\\\n=0"


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