(a) Explanation Given: A single heat source located at the origin with temperature T ( x , y , z ) = 25 x 2 + y 2 + z 2 , and the surface given by S = { ( x , y , z ) : z = 1 − x 2 , − 1 2 ≤ x ≤ 1 2 , 0 ≤ y ≤ 1 } Refer to the question for figure. Formula used: ∂ ∂ x x n = n x n − 1 , ∂ ∂ y y n = n y n − 1 , ∂ ∂ z z n = n z n − 1 ( f g ) ′ = g f ′ − f g ′ g 2 Calculation: The heat flux across the surface S is given by − K ∬ S ∇ T ( x , y , z ) . N d S , where K is the conductivity constant. We have, T ( x , y , z ) = 25 x 2 + y 2 + z 2 , such that ∇ T ( x , y , z ) = ∇ ( 25 x 2 + y 2 + z 2 ) ∇ T ( x , y , z ) = ( ∂ ∂ x i ^ + ∂ ∂ y j ^ + ∂ ∂ z k ^ ) ( 25 x 2 + y 2 + z 2 ) ∇ T ( x , y , z ) = ∂ ∂ x ( 25 x 2 + y 2 + z 2 ) i ^ + ∂ ∂ y ( 25 x 2 + y 2 + z 2 ) j ^ + ∂ ∂ z ( 25 x 2 + y 2 + z 2 ) k ^ ∇ T ( x , y , z ) = 25 ∂ ∂ x ( x 2 + y 2 + z 2 ) − 1 2 i ^ + ∂ ∂ y ( x 2 + y 2 + z 2 ) − 1 2 j ^ + ∂ ∂ z ( x 2 + y 2 + z 2 ) − 1 2 k ^ Applying the formula ∂ ∂ x x n = n x n − 1 , ∂ ∂ y y n = n y n − 1 , ∂ ∂ z z n = n z n − 1 and the Quotient rule ( f g ) ′ = g f ′ − f g ′ g 2 , where f and g are both differentiable ∇ T ( x , y , z ) = − 25 2 [ ( x 2 + y 2 + z 2 ) − 3 2 ( 2 x ) i ^ + ( x 2 + y 2 + z 2 ) − 3 2 ( 2 y ) j ^ + ( x 2 + y 2 + z 2 ) − 3 2 ( 2 z ) k ^ ] ∇ T ( x , y , z ) = − 25 [ ( x 2 + y 2 + z 2 ) − 3 2 ( x ) i ^ + ( x 2 + y 2 + z 2 ) − 3 2 ( y ) j ^ + ( x 2 + y 2 + z 2 ) − 3 2 ( z ) k ^ ] ∇ T ( x , y , z ) = − 25 x i ^ + y j ^ + z k ^ ( x 2 + y 2 + z 2 ) 3 2 For the surface given by S = { ( x , y , z ) : z = 1 − x 2 , − 1 2 ≤ x ≤ 1 2 , 0 ≤ y ≤ 1 } , projecting it onto the xy-plane, such that g ( x , y ) = z = 1 − x 2 Now, partially differentiating g ( x , y ) = z = 1 − x 2 with respect to ‘ x ’ ∂ ∂ x g ( x , y ) = ∂ ∂ x 1 − x 2 g x ( x , y ) = ∂ ∂ x ( 1 − x 2 ) 1 2 g x ( x , y ) = 1 2 ( 1 − x 2 ) − 1 2 ∂ ∂ x ( 1 − x 2 ) g x ( x , y ) = 1 2 ( 1 − x 2 ) − 1 2 ( − 2 x ) g x ( x , y ) = − x 1 − x 2 partially differentiating g ( x , y ) = z = 1 − x 2 with respect to ‘ y ’ ∂ ∂ y g ( x , y ) = ∂ ∂ y 1 − x 2 g y ( x , y ) = 0 If N be the outward normal vector to the given surface S, then we can use the fact that, N d S = − g x i − g y j + k 1 + [ g x ] 2 + [ g y ] 2 1 + [ g x ] 2 + [ g y ] 2 d A N d S = ( − g x i − g y j + k ) d A N d S = ( x 1 − x 2 i ^ + k ^ ) d A Heat Flux = − K ∬ S ∇ T ( x , y , z ) (b) To determine To Calculate: the heat flux across the surface S using the given parametrization

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Author: Ron Larson; Bruce H. Edwards

Publisher: Cengage Limited

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Chapter 15, Problem 1PS

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To Calculate: the heat flux across the surface S

(b)

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To Calculate: the heat flux across the surface S using the given parametrization

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the heat flux across the surface S

Solution Summary: The author explains how the heat flux across the surface S is given by 25pi K3sqrt2.

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