15_6_7 - is ZZ F • ˆ N dS = -ZZ D ( 2 xy 3 + 2 y ( 4-x...

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7. The part of z = 4 - x 2 - y 2 lying above z = 2 x + 1 has projection onto the xy -plane the disk D bounded by 2 x + 1 = 4 - x 2 - y 2 , or ( x + 1 ) 2 + y 2 = 4 . Note that D has area 4 π and centroid ( - 1 , 0 ) . For z = 4 - x 2 - y 2 , the downward vector surface element is ˆ N dS = - 2 x i - 2 y j - k 1 dx dy . Thus the flux of F = y 3 i + z 2 j + x k downward through
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Unformatted text preview: is ZZ F • ˆ N dS = -ZZ D ( 2 xy 3 + 2 y ( 4-x 2-y 2 ) 2 + x ) dx dy (use the symmetry of D about the x-axis) = -ZZ D x d A = -( 4 π)(-1 ) = 4 π....
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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