Seawater has density 1025 kg/m3 and flows in a velocity field v = y i + x j, where x, y, and z are measured in meters and the components of v in meters per second. find the rate of flow outward through the hemisphere x2 + y2 + z2 = 81, z ≥ 0.

Flow (Q) = 0 kg/s      

Step-by-step explanation:

Given:-

- The density of seawater, ρ = 1025 kg/m^3

- The velocity field of flow, v ( x,y,z ) = y i + x j

Find:-

The rate of flow outward through the hemisphere x2 + y2 + z2 = 81, z ≥ 0.

Solution:-

For some flow, if we know the point density at (x,y,z) is ρ ( x , y ,z ).

And the velocity of a unit element at point (x,y,z) is v ( x , y ,z ).

Then the rate of outflow of substance through surface "S" is given by:

Where,                    F = ρ ( x , y ,z )*v ( x , y ,z )

- Therefore for the given data:

                               F = ( 1025 kg/m^3)*(y i + x j)

- We will now parametrize the surface "S" given as:

                              x2 + y2 + z2 = 81, z ≥ 0

           S: r(u) = 3 sin (v) cos (u) i + 3 sin (v) sin (u) j + 3 cos (v) k  

Where,         u ε [ 0 , 2π) and since z ≥ 0, v ε [ 0 , π/2 ]

Therefore,

                   F [ r(u) ] = ( 1025 kg/m^3)*( 3 sin (v) sin (u) i + 3 sin (v) cos (u) j).

Note:

Where,           n: Unit vector normal to surface "S"

- Such that,          

The Flux through a surface is defined only if the surface is orientable.

Note that if the substance is orientable, there will be two normal unit vectors at every point (x,y,z):

- We can write:

- For finding the flux we always use the positive orientation of unit normal vector (n).

- The parameterisation for sphere "S":

              r ( u , v ) = 3 sin (v) cos (u) i + 3 sin (v) sin (u) j + 3 cos (v) k          

Where,         u ε ( 0 , 2π) and since z ≥ 0, v ε ( 0 , π/2 ),

- Compute ( r_u x r_v ) :

- The vector ( r_u x r_v) points towards the origin, therefore the positive unit normal vector ( r_u x r_v) would be:

Therefore,

So the flow through the surface "S" is Flow (Q) = 0 kg/s            

the distance from any point on the parabola to the focus is the exact same as the distance from that point to the directrix.

if you could provide answer choices that might allow me to answer the question easier if my above answer isn't an option! (: