15_2_18 - rate 2 π per unit length along the line. Thus,...

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18. The velocity field for a point source of strength m dt at ( 0 , 0 , t ) is v t ( x , y , z ) = m ( x i + y j + ( z - t ) k ) ( x 2 + y 2 + ( z - t ) 2 ) 3 / 2 . Hence we have Z -∞ v t ( x , y , z ) dt = m Z -∞ x i + y j + ( z - t ) k ( x 2 + y 2 + ( z - t ) 2 ) 3 / 2 dt = m ( x i + y j ) Z -∞ dt ( x 2 + y 2 + ( z - t ) 2 ) 3 / 2 Let z - t = p x 2 + y 2 tan θ - dt = p x 2 + y 2 sec 2 θ d θ = m ( x i + y j ) x 2 + y 2 Z π/ 2 - π/ 2 cos θ d θ = 2 m ( x i + y j ) x 2 + y 2 , which is the velocity field of a line source of strength 2 m along the z -axis. The definition of strength of a point source in 3-space was made to ensure that the velocity field of a source of strength 1 had speed 1 at distance 1 from the source. This corresponds to fluid being emitted from the source at a volume rate of 4 π . Similarly, the definition of strength of a line source guaranteed that a source of strength 1 gives rise to fluid speed of 1 at unit distance 1 from the line source. This corresponds to a fluid emission at a volume
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Unformatted text preview: rate 2 π per unit length along the line. Thus, the integral of a 3-dimensional source gives twice the volume rate of a 2-dimensional source, per unit length along the line. The potential of a point source m dt at ( , , t ) is φ( x , y , z ) = -m p x 2 + y 2 + ( x-t ) 2 . This potential cannot be integrated to give the potential for a line source along the z-axis because the integral-m Z ∞-∞ dt p x 2 + y 2 + ( z-t ) 2 does not converge, in the usual sense in which convergence of improper integrals was defined....
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