quiz_08D - y ), therefore RR D x 2 y 3 dA = 0 since D is...

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Math 213 — Quiz 8 Name Solution 1. Use symmetry to evaluate RR D ( x 2 tan( x ) + y 3 + 4) dA where D = { ( x,y ) | x 2 + y 2 2 } . First note that ZZ D ( x 2 tan( x ) + y 3 + 4) dA = ZZ D x 2 tan( x ) dA + ZZ D y 3 dA + ZZ D 4 dA. The function x 2 tan( x ) has odd symmetry in x (i.e. ( - x ) 2 tan( - x ) = - ( x 2 tan( x )) for every x ), therefore RR D x 2 tan( x ) dA = 0 since D is also symmetric in x (i.e. ( x,y ) D ( - x,y ) D ). The function y 3 has odd symmetry in y (i.e. ( - y ) 3 = - ( y 3 ) for every
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Unformatted text preview: y ), therefore RR D x 2 y 3 dA = 0 since D is also symmetric in y (i.e. ( x,y ) D ( x,-y ) D ). Finally, RR D 4 dA = 8 since this integral is the volume of a cylinder of radius 2 and height 4. Therefore RR D ( x 2 tan( x ) + y 3 + 4) dA = 8 ....
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This note was uploaded on 05/09/2008 for the course MATH 2130 taught by Professor Dorais during the Spring '08 term at Cornell University (Engineering School).

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