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Unformatted text preview: part 1. Solution: The exact value of the given integral is Z & Z & sin ( x + y ) dy dx . Evaluating the inner integral, we obtain Z & sin ( x + y ) dy = & cos ( x + y ) j y = & y =0 = & cos ( x + & ) + cos ( x ) = & (cos ( x ) cos ( & ) & sin ( x ) sin ( & )) + cos ( x ) = 2 cos ( x ) . This gives Z & Z & sin ( x + y ) dy dx = Z & 2 cos ( x ) dx = 2 sin ( x ) j x = & x =0 = 0 . Here is a graph of f ( x; y ) = sin ( x + y ) . Note that the equal parts of the graph lie above and below the plane z = 0 . This explains why the integral is equal to . 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