# Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38 866 SECTION 17.1 CHAPTER 17 SECTION 17.1 1. 3 i =1 3 j =1 2 i 1 3 j +1 = 3 i =1 2 i 1 3 j =1 3 j +1 = (1 + 2 + 4)(9 + 27 + 81) = 819 2. 2 + 2 2 + 3 + 3 2 + 4 + 4 2 + 5 + 5 2 = 68 3. 4 i =1 3 j =1 ( i 2 + 3 i )( j 2) = 4 i =1 ( i 2 + 3 i ) 3 j =1 ( j 2) = (4 + 10 + 18 + 28)( 1 + 0 + 1) = 0 4. 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 4 2 + 4 3 + 4 4 + 4 5 + 4 6 + 4 7 + 6 2 + 6 3 + 6 4 + 6 5 + 6 6 + 6 7 = 19 4 35 . 5. m i =1 Δ x i = Δ x 1 + Δ x 2 + · · · + Δ x m = ( x 1 x 0 ) + ( x 2 x 1 ) + · · · + ( x m x m 1 ) = x m x 0 = a 2 a 1 6. ( y 1 y 0 ) + ( y 2 y 1 ) + · · · + ( y n y n 1 ) = y n y 0 = b 2 b 1 7. m i =1 n j =1 Δ x i Δ y j = m i =1 Δ x i n j =1 Δ y j = ( a 2 a 1 )( b 2 b 1 ) 8. n j =1 q k =1 Δ y j Δ z k = n j =1 Δ y j q k =1 Δ z k = ( b 2 b 1 )( c 2 c 1 ) 9. m i =1 ( x i + x i 1 x i = m i =1 ( x i + x i 1 )( x i x i 1 ) = m i =1 ( x i 2 x 2 i 1 ) = x m 2 x 0 2 = a 2 2 a 1 2 10. n j =1 1 2 ( y j 2 + y j y j 1 + y j 1 2 y j = 1 2 n j =1 ( y j 3 y j 1 3 ) = 1 2 ( b 2 3 b 1 3 ) 11. m i =1 n j =1 ( x i + x i 1 x i Δ y j = m i =1 ( x i + x i 1 x i n j =1 Δ y j (Exercise 9) = ( a 2 2 a 1 2 ) ( b 2 b 1 ) 12. m i =1 n j =1 ( y i + y j 1 x i Δ y j = m i =1 Δ x i n j =1 ( y j 2 y j 1 2 ) = ( a 2 a 1 )( b 2 2 b 1 2 )

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38 SECTION 17.2 867 13. m i =1 n j =1 (2Δ x i y j ) = 2 m i =1 Δ x i n j =1 1 3 m i =1 1 n j =1 Δ y j = 2 n ( a 2 a 1 ) 3 m ( b 2 b 1 ) 14. m i =1 n j =1 (3Δ x i y j ) = 3 m i =1 n j =1 Δ x i 2 m i =1 n j =1 Δ y j = 3 n ( a 2 a 1 ) 2 m ( b 2 b 1 ). 15. m i =1 n j =1 q k =1 Δ x i Δ y j Δ z k = m i =1 Δ x i n j =1 Δ y j q k =1 Δ z k = ( a 2 a 1 )( b 2 b 1 )( c 2 c 1 ) 16. m i =1 n j =1 q k =1 ( x i + x i 1 x i Δ y j Δ z k = m i =1 ( x i 2 x i 1 2 ) n j =1 Δ y j q k =1 Δ z k = ( a 2 2 a 1 2 )( b 2 b 1 )( c 2 c 1 ) 17. n i =1 n j =1 n k =1 δ ijk a ijk = a 111 + a 222 + · · · + a nnn = n p =1 a ppp 18. Start with m i =1 n j =1 a ij . Take all the a ij (there are only a finite number of them) and order them in any order you chose. Call the first one b 1 , the second b 2 , and so on. Then m i =1 n j =1 a ij = r p =1 b p where r = m × n. SECTION 17.2 1. L f ( P ) = 2 1 4 , U f ( P ) = 5 3 4 2. L f ( P ) = 3 , U f ( P ) = 5 3. (a) L f ( P ) = m i =1 n j =1 ( x i 1 + 2 y j 1 x i Δ y j , U f ( P ) = m i =1 n j =1 ( x i + 2 y j x i Δ y j (b) L f ( P ) m i =1 n j =1 x i 1 + x i 2 + 2 y j 1 + y j 2 Δ x i Δ y j U f ( P ) . The middle expression can be written m i =1 n j =1 1 2 ( x i 2 x 2 i 1 ) Δ y j + m i =1 n j =1 ( y j 2 y 2 j 1 ) Δ x i . The first double sum reduces to m i =1 n j =1 1 2 ( x i 2 x 2 i 1 ) Δ y j = 1 2 m i =1 ( x i 2 x 2 i 1 ) n j =1 Δ y j = 1 2 (4 0)(1 0) = 2 . In like manner the second double sum also reduces to 2 . Thus, I = 4; the volume of the prism bounded above by the plane z = x + 2 y and below by R .
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38 868 SECTION 17.2 4. L f ( P ) = 7 / 16 , U f ( P ) = 7 / 16 5. L f ( P ) = 7 / 24 , U f ( P ) = 7 / 24 6. (a) L f ( p ) = m i =1 n j =1 ( x i 1 y j

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