tutorial6_sol

tutorial6_sol - R b a R d c M dy dx Here LHS is volume...

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MATH1211/10-11(2)/Tu6sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 6 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. (a) We deﬁne D ( a,b,c ) = n i =1 ( y i - ( ax 2 i + bx i + c ) ) 2 . (b) The required system of equations is ∂D ∂a = ∂D ∂b = ∂D ∂c = 0, which is ∑( y i - ( ax 2 i + bx i + c ) ) ( - x 2 i ) = 0 ∑( y i - ( ax 2 i + bx i + c ) ) ( - x i ) = 0 ∑( y i - ( ax 2 i + bx i + c ) ) ( - 1) = 0 . Here the sums are from i = 1 to i = n . The equations reduce to the system of linear equations (∑ x 4 i ) a + (∑ x 3 i ) b + (∑ x 2 i ) c = x 2 i y i (∑ x 3 i ) a + (∑ x 2 i ) b + ( x i ) c = x i y i (∑ x 2 i ) a + ( x i ) b + n c = y i . 2. The answer is Z 2 1 Z 3 0 ( x + 3 y + 1) dxdy = Z 2 1 h x 2 2 + x (3 y + 1) i x =3 x =0 dy = Z 2 1 ± 9 y + 15 2 ² dy = 9 y 2 2 + 15 y 2 ³ ³ ³ 2 1 = 21 . You may also set up the iterated integral as R 3 0 R 2 1 ( x +3 y +1) dy dx to compute the same numerical answer. 3. Because f M , we have R b a R d c f ( x,y ) dy dx
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Unformatted text preview: R b a R d c M dy dx . Here LHS is volume under the graph of f over R , and RHS is R b a ( My ) ³ ³ d c dx = R b a M ( d-c ) dx = M ( d-c ) x ³ ³ b a = M ( b-a )( d-c ). The result then follows. 4. The bounded region is shown as in the ﬁgure at right. Intersecting points of the two curves are (0 , 0) and ( 1 4 , 1 2 ) , with y = √ x being the upper curve and y = 32 x 3 the lower curve that bound the region. Hence the integral is evaluated as ZZ R 3 xy dA = Z 1 4 Z √ x 32 x 3 3 xy dy dx = Z 1 4 ± 3 xy 2 2 ²³ ³ ³ ³ y = √ x y =32 x 3 dx = Z 1 4 ± 3 x 2 2-1536 x 7 ² dx = ± x 3 2-192 x 8 ²³ ³ ³ ³ 1 4 = 1 128-3 1024 = 5 1024 . 1...
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