Question

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.

Answer 1

Answer:

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:

                                $Flow (Q) = \int\int\limits_S {F} \, .ds$

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:

                       $Flow (Q) = \int\int\limits_S {F} \, .ds = \int\int\limits_D {F} \, .n.ds$

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

- Such that,          $dS = magnitude(r_u x r_v).dA$

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:

               $dS = n.dA = +/- \frac{r_u x r_v}{magnitude (r_u x r_v)}*magnitude (r_u x r_v).dA \\\\dS = +/- (r_u x r_v).dA$                  

- 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 ),

              $r_u = -3 sin(v)*sin(u) i + 3 sin(v)*cos (u) j \\\\r_v = 3 cos (v) *cos (u) i + 3 cos (v)* sin (u) j - 3 sin (v) k$

- Compute ( r_u x r_v ) :

             $r_u xr_v = \left[\begin{array}{ccc}i&j&k\\-3sin(v)*sin(u)&3sin(v)*cos(u)&0\\3cos(v)*cos(u)&3cos(v)*sin(u)&3sin(v)\end{array}\right] \\\\\\r_u xr_v = -9 sin^2(v)*cos(u) i -9 sin^2(v)*sin(u)j -9 sin(v)*cos(v) k$

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

$dS = - (r_u xr_v).dA = [ 9 sin^2(v)*cos(u) i +9 sin^2(v)*sin(u)j +9 sin(v)*cos(v) k] .dA$

Therefore,

               $Flow (Q) = \int\int\limits_S {F} \, .ds = 3075\int\int\limits_D { [2sin^3(v)*sin(u)*cos(u) ]} . dA\\\\Flow (Q) = 3075\int\int\limits_D { [sin^3(v)*sin(2u) ]} . dA\\\\\\u = e ( a1 = 0 , b1 = 2\pi ), v = e ( a2 = 0 , b2 = \pi/2)\\\\\\Flow (Q) = 3075\int\limits^b_0 {sin(2u)} \, du*\int\limits^b_0 {sin^3(v)} \, dv\\\\Flow (Q) = 3075[ - \frac{cos(2u)}{2}]\limits^2^\pi _0 * \int\limits^b_0 {sin^3(v)} \, dv\\\\Flow (Q) = 3075[ 0 ] * \int\limits^b_0 {sin^3(v)} \, dv = 0$

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