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Unformatted text preview: part 1. Solution: The exact value of the given integral is Z & Z & cos ( x + y ) dy dx . Evaluating the inner integral, we obtain Z & cos ( x + y ) dy = sin ( x + y ) j y = & y =0 = sin ( x + & ) & sin ( x ) = sin ( x ) cos ( & ) + cos ( x ) sin ( & ) & sin ( x ) = & 2 sin ( x ) . This gives Z & Z & cos ( x + y ) dy dx = & Z & 2 sin ( x ) dx = 2 cos ( x ) j x = & x =0 = 2 cos ( & ) & 2 cos (0) = & 4 . Here is a picture of f ( x; y ) = cos ( x + y ) . Note that most of the graph lies below the plane z = 0 , which explains why the integral is a negative number. 2 3...
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This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.
- Fall '10