document.write( "Question 1062026: Compute for the value of \r
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Algebra.Com's Answer #677069 by ikleyn(52781)\"\" \"About 
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document.write( "1.  Let me show first how to calculate more simple integral \r\n" );
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document.write( "    .      (1)\r\n" );
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document.write( "    Change variables for polar coordinates x = \"r%2Acos%28phi%29\",  y = \"r%2Asin%28phi%29\".\r\n" );
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document.write( "    Then dx*dy = \"r%2Adr%2Ad%28phi%29\", and the integral (1) becomes \r\n" );
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document.write( "     =  = (introduce new variable z = \"r%5E2\") = .\r\n" );
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document.write( "    The internal integral is \r\n" );
document.write( "                                                          |\"infinity\"\r\n" );
document.write( "    \"int%28%281%2F%281%2B4z%29%5E2%29%2C+dz%2C0%2C+infinity%29\" = \"%281%2F4%29%2A%28int%28%281%2F%281%2B4z%29%5E2%29%2C+d%284z%29%2C0%2C+infinity%29%29%29\"  = \"%28-1%2F4%29%2A%281%2F%281+%2B+4z%29%29\" |         =  \"1%2F4\".\r\n" );
document.write( "                                                          |0\r\n" );
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document.write( "    Then the entire double integral (1) is  \"%281%2F2%29%2A%281%2F4%29%2A2pi\" = \"pi%2F4\".\r\n" );
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document.write( "2.  Now, let us start with the original integral  \r\n" );
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document.write( "    .      (2)\r\n" );
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document.write( "To calculate this integral, let me introduce the new coordinate system \r\n" );
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document.write( "   u = x,\r\n" );
document.write( "   v = \"%283%2F2%29%2Ay\".\r\n" );
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document.write( "Then\r\n" );
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document.write( "   x = u,  y = \"%282%2F3%29%2Av\";  dx = du, dy = \"%282%2F3%29%2Adv\", and the integral (2) becomes\r\n" );
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document.write( "   .\r\n" );
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document.write( "   The integral of \"u\" and \"v\" after the factor \"2%2F3%29\" was just calculated in the section #1.\r\n" );
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document.write( "   Hence, the final answer is \"%282%2F3%29%2A%28pi%2F4%29\" = \"pi%2F6\".\r\n" );
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