In R3 filled with an incompressible fluid, we say that the z-axis is a line source ofstrength m if every interval ∆z along that axis emits fluid at volume rate dV /dt = 2πm ∆z. The fluidthen spreads out symmetrically in all directions perpendicular to the z-axis. Show that the velocityfield of the flow isv(x, y) = mx2 + y2 (xi + yj).

The problem is asking to show that the velocity field of the flow is v(x, y) = m/(x^2 + y^2) (xi + yj). This is a problem of fluid dynamics, specifically dealing with the concept of a line source in an incompressible fluid.

Here are the steps to solve this problem:

1. We start with the given that the volume rate of fluid emission is dV/dt = 2πm Δz. This is the rate at which fluid is being emitted from an interval Δz along the z-axis.

2. The fluid spreads out symmetrically in all directions perpendicular to the z-axis. This means that at any point (x, y) in the plane perpendicular to the z-axis, the velocity of the fluid is the same.

3. The velocity field of the flow, v(x, y), is a vector field that gives the velocity of the fluid at every point in the plane. We want to show that this is given by v(x, y) = m/(x^2 + y^2) (xi + yj).

4. The magnitude of the velocity at a point (x, y) is given by the volume rate divided by the area over which the fluid is spreading. The area of a circle with radius r is πr^2, and in this case r = sqrt(x^2 + y^2). So the magnitude of the velocity is (2πm Δz) / (π(x^2 + y^2) Δz) = 2m / (x^2 + y^2).

5. The direction of the velocity at a point (x, y) is along the radial line from the origin to that point. In terms of the unit vectors i and j in the x and y directions, this direction is (x/sqrt(x^2 + y^2))i + (y/sqrt(x^2 + y^2))j = xi + yj.

6. Therefore, the velocity field of the flow is v(x, y) = m/(x^2 + y^2) (xi + yj), as required.