237t2s - 1. Let the surface S be parametrized by f (u , v )...

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1. Let the surface S be parametrized by ) , , 2 ( ) , ( 2 v u e u v u v u v + + + = f and let be a point in ) 1 , 2 , 2 ( 0 P 3 R . (a) [2 marks] Show that the point lies on the surface S . 0 P We should be able to find ( u , v ) such that ) 1 , 2 , 2 ( ) , ( = v u f . We have the system of equations 1 , 2 , 2 2 2 = + = + = + v u e u v u v . From the second equation we get since on the right hand side we do not have e and also u =1, to satisfy that equation. 0 = v Its easy to see that the other two equations are satisfied also. So lies on the surface S 0 P and it corresponds to ) 0 , 1 ( ) , ( = v u (b) [5 marks] Write the equation for the tangent plane to S at the point in the 0 P form d cz by ax = + + , where are constants. d c b a , , , Vector normal to the surface S at the point is 0 P ) 0 , 1 )( ( v u × = f f N , ) 2 , 1 , 1 ( ) 2 , 1 , 1 ( ) 0 , 1 )( ( 0 1 = + = = = v u u v u u f , ) 1 , 1 , 1 ( ) 1 , , 1 ( ) 0 , 1 )( ( 0 1 = + = = = v u v e v u v f . So N = = ) 1 , 1 , 1 ( ) 2 , 1 , 1 ( × ) 2 , 1 , 3 ( and the equation of the tangent plane to S at the point is 0 P 0 ) 1 )( 2 ( ) 2 ( 1 ) 2 ( 3 = + + z y x or 6 2 3 = + z y x . 2
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2. Let V be the set in the first octant in 3 R bounded by the surface 8 4 2 3 3 3 = + + z y x and the coordinate planes and let the map be defined by . 3 3 : R R G ) , , ( ) , , ( 3 3 3 w v u w v u = G (a) [7 marks] Apply the change of variables ) , , ( ) , , ( w v u z y x G = to the integral . Write the iterated integral for I in the new variables in ∫∫∫ + + = V dV z y x I ) 1 ln( such an order of integration that there are NO FRACTIONS in the limits of integration. DO NOT EVALUATE THE INTEGRAL We have to describe such that . It is easily seen that is the tetrahedron with vertices at (4, 0, 0), (0, 8, 0), (0, 0, 2). To avoid fractions we have to project on uw -plane and let . On uw -plane we have the region bounded by coordinate axes and the line . Hence we can describe it as V ) ( = V V G V V w u v 4 2 8 0 4 2 = + w u w u 2 4 0 , 2 0 w . So describe as V follows: } 2 0 , 2 4 0 , 4 2 8 0 : ) , , {( = w w u w u v w v u V . Given function s continuous, hence integrable and we get R R V f 3 : i dudvdw w v u w v u dV z y x I V V ) , , ( det )] 1 ln( [ ) 1 ln( 3 3 3 G + + = + + = ∫∫∫ ∫∫∫ . Now , and in the required order of integration, the iterated integral can be written as 2 2 2 27 ) ( det w v u u = G dw du dv w v w v u I ww u ) 1 ln( 27 3 2 0 2 4 0 4 2 8 0 3 2 2 5 + + = ∫∫ ∫ −− (b) [2 marks] Explain briefly , why the given change of variables could be applied for I although 0 ) , , ( det = w v u D G on some subset of the inverse image of V under G . The applicability of given change of variables follows directly from “The Change of Variables Theorem”. The map G is of class on 1 C 3 R and on any open box ) , 0 ( ) , 0 ( ) , 0 ( 0 a a a U × × = , with a > 8, G is 1-1 on and 0 U 0 ) , , ( det w v u D G on . The open set 0 U } 2 0 , 2 4 0 , 4 2 8 0 : ) , , {( < < < < < < = w w u w u v w v u T
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This note was uploaded on 10/21/2010 for the course MATHEMATIC MAT237Y1 taught by Professor Romauldstanczak during the Fall '09 term at University of Toronto- Toronto.

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237t2s - 1. Let the surface S be parametrized by f (u , v )...

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