Tutorial 10 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1505 Mathematics I Tutorial 10 (Solution Notes) 1. Evaluate ± S f ( x, y, z ) dS and ± S F dS ,where f ( x, y, z )= x + y + z and F = x 2 i + y 2 j + z 2 k and S is the surface deFned parametrically by r ( u, v )=(2 u + v ) i +( u 2 v ) j u +3 v ) k , (0 u 1 , 0 v 2) . The orientation of S is given by the normal vector r u × r v . Preliminaries Parametric Equation The parametric equation of a curve in 3-dimensional space is of the form: x ( t f 1 ( t ) ,y ( t f 2 ( t ) ,z ( t f 3 ( t ) , or written as r ( t f 1 ( t ) i + f 2 ( t ) j + f 3 ( t ) k . The parametric equation of a surface in 3-dimensional space is of the form: x ( u, v f 1 ( u, v ) ( u, v f 2 ( u, v ) ( u, v f 3 ( u, v ) , or written as r ( u, v f 1 ( u, v ) i + f 2 ( u, v ) j + f 3 ( u, v ) k . Remark: 1. It is quite natural to use two parameters in the equations for surface. 2. You should be familiar with some special kinds of surface. 3. Try to Fnd the parametric equations of the exercises in section 12.6 of Thomas’ Calculus. Surface Integral Same as line integrals, we still have two types of formulas that can directly compute the surface integral: For line integral, we use r ± ( t ) , which is the direction of the line. For surface integral, we use the cross product of two partial derivatives r u × r v ,wh ichis the normal vector of the tangent plane. 33 Check details in lecture notes, section 10.1.5. 33 r u and r v are two vectors contained in the tangent plane, and so the cross product gives its normal vector. 198
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One is for scalar Felds (scalar functions): ±± S f ( x, y, z ) dS = R f ( r ( u, v )) || r u × r v || dA. The other is for vector Felds: S F ( x, y, z ) d S = R F ( r ( u, v )) ( r u × r v ) dA. Remark: 1. The formulas are similar to the formulas of line integral: for scalar functions, we multiply the magnitude; for the vector Felds, we use dot product. 2. Given a function (Feld), we can easily determine which formula to use. 3. Usually, two difficulties to apply these formulas: Fnd out the expression of the function, and Fnd out the parametric equations of the surface. Solution The solution is quite obvious. Since we need to apply the two formulas for surface integrals listed above, so the solution steps are: Find f ( r ( u, v )) , F ( r ( u, v )) and r u × r v . Substitute them into the formula, and estimate the double integral on the right hand side (sketch the region R , and change the double integral to iterated integral). ±rom the conditions given in the question, we know f ( x, y, z )= x + y + z and F ( x, y, z x 2 i + y 2 j + z 2 k , so substitute the parametric equation of the surface r ( u, v )=(2 u + v ) i +( u 2 v ) j u +3 v ) k into the two functions, that is, we substitute x =2 u + v, y = u 2 z = u v and get: f ( r ( u, v )) = f (2 u + v, u 2 v ) =( 2 u + v )+( u 2 v u v ) =4 u +2 v. F ( r ( u, v )) = F (2 u + 2 v ) 2 u + v ) 2 i u 2 v ) 2 j u v ) 2 k . 199
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Then we take partial derivatives of the parametric equation r ( u, v ): r u = ∂u (2 u + v ) i + ( u 2 v ) j + ( u +3 v ) k =2 i + j + k r v = ∂v (2 u + v ) i + ( u 2 v ) j + ( u v ) k = i 2 j k and so, r u × r v =( 2 i + j + k ) × ( i 2 j k ) = ± ± ± ± ± ± ± ijk 211 1 23 ± ± ± ± ± ± ± = ² 1 × 3 ( 2) × 1 ³ i ² 2 × 3 1 × 1
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Tutorial 10 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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