notes001 (28)

notes001 (28) - part 1. Solution: The exact value of the...

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MATH 2203 - Quiz 5 (Version 1) Solution November 2, 2009 NAME______________________ 1. (a) Use a Riemann sum with m = n = 2 to estimate the value of ZZ R sin ( x + y ) dA where R = [0 ] [0 ] . Take the sample points to be the lower left corners. Solution: We have x = ± 0 2 = 2 and likewise y = 2 . Therefore A x y = 2 4 . The lower left corners of the subrectangles are (0 ; 0) ; ( 2 ; 0) ; (0 2) ; and ( 2 2) . Therefore the Riemann sum estimate is 2 4 sin (0 + 0) + sin 2 + 0 ± + sin 0 + 2 ± + sin 2 + 2 ±± = 2 4 (2) = 2 2 . (b) Obtain another estimate for this integral by using the midpoints as sample points. Solution: The midpoints of the subrectangles are ( 4 4) ; (3 4 4) ; ( 4 ; 3 4) ; and (3 4 ; 3 4) . Therefore the Rie- mann sum estimate is 2 4 ² sin 4 + 4 ± + sin ² 3 4 + 4 ³ + sin ² 4 + 3 4 ³ + sin ² 3 4 + 3 4 ³³ = 0 . 1
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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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notes001 (28) - part 1. Solution: The exact value of the...

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