MATH
260S12Ex3solns

# Parametrize γ as γ t t 2 t 3 t for 0 t 1 then γ t

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Parametrize γ as γ ( t ) = ( t, 2 t, - 3 t ), for 0 t 1. Then γ ( t ) = (1 , 2 , - 3) and on γ we have F ( γ ( t )) · dt = (2 t, 3 + 2 t, 2) · (1 , 2 , - 3) = 2 t + 2(3 + 2 t ) - 6 = 6 t. Therefore γ F · d s = 1 0 6 t dt = 3 . B–2. Let G ( x ) := b ( x ) a ( x ) f ( t ) dt , where a ( x ) and b ( x ) are smooth functions with a ( x ) < b ( x ), and f ( x ) is a continuous function. Compute dG ( x ) /dx . Solution: For real numbers p < q let H ( p, q ) := q p f ( t ) dt , Then G ( x ) = H ( a ( x ) , b ( x )) so by the chain rule with p = a ( x ), and q = b ( x ) dG dx = ∂H ∂p dp dx + ∂H ∂q dq dx = - f ( a ( x )) da dx + f ( b ( x )) db dx = f ( b ( x )) db dx - f ( a ( x )) da dx . 2

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B–3. Compute J := R 2 dx dy [4 + ( x - y ) 2 + ( x + 2 y ) 2 ] 2 Solution: Since we recognize that R 2 du dv [4 + u 2 + v 2 ] 2 is probably doable by using polar coordinates, it is natural to try the preliminary substitution u = x - y , v = x + 2 y . Then ( u, v ) ( x, y ) = det 1 - 1 1 2 = 3 so du dv = 3 dx dy. Thus, J = R 2 1 [4 + u 2 + v 2 ] 2 du dv 3 . Now we change to polar coordinates, u = r cos θ , v = r sin θ and find J = 1 3 2 π 0 0 1 [4 + r 2 ] 2 r dr = π 12 B–4. Let the surface S R 3 be the graph of z = g ( x, y ) for ( x, y ) in a region D in the xy - plane. a) Using the parameters x = u , y = v , z = g ( u, v ), derive the formula Area ( S ) = D 1 + g 2 dx dy. Solution: We begin with the formula for the area of a surface defined parametrically over a region Ω in the ( u, v ) parameter space: Area ( S ) = Ω T u × T v du dv, where T u = ( x u , y u ) and T v = ( x v , y v ) are tangent vectors on the surface. In this special case, T u = (1 , 0 , g u ( u, v ) and T v = (0 , 1 , g v ( u, v ). Then by a routine computation, T u × T v = ( - g u , - g v , 1) so T u × T v = 1 + g 2 u + g 2 v . Since
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