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By Snieder R.

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Problem d: Show that V = R v dS (where the integration is over an arbitrary surface around the source at r = 0). 9) v(r) = V ^r : 2 r From this simple example of a single source at r = 0 more complex examples can be obtained. Suppose we have a source at r+ = (L 0) where a volume V is injected per unit time and a sink at r; = ( L 0) where a volume V is removed per unit time. 9) for the source and the sink. 11) and sketch the resulting ow eld. This is most easily accomplished by determining from the expressions above the ow eld at some selected lines such as the x- and y-axes.

The reason that the vorticity vanishes is that the contribution of the rotation around the z -axis to the vorticity is equal but of opposite sign from the contribution of the shear, so that the total vorticity vanishes. Note that this implies that a paddle-wheel does not change its orientation as it moves with this ow! 5. 11) is contrived in an arti cial way. However, keep in mind that all the arguments of the previous section apply to any vector eld and that uid ow was used only as an example to x our mind.

F1 = 0). Let us simplify the notation further by dropping the subscript "1" in p1 . 11) 0 0 r S ; 0 r This result is called the "representation theorem" because it gives the wave eld inside the volume when the wave eld (an its gradient) are speci ed on the surface that bounds this volume. 11) can be used to formally derive Huygens' principle which states that every point on a wavefront acts as a source for other waves and that interference of these waves determine the propagation of the wavefront.

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