Consider the following dimensionless scalar field
$$S = x + y + zt$$
Apply the Reynolds Transport Theorem to evaluate the volume integral over the space specified by the surfaces
$$x = 0\; \& \; t,\;\;\;\;\; y = 0\ \& \ 2t,\;\;\;\; z = 0\; \&\; 4t$$
I am stuck with this integral, $$\int_S S (V_S∙n ̂) dS $$
How to integrate a scalar field over surfaces that depend on time ?
Can anybody help me with this ?
Reynolds Transport Theorem: $$ \frac{D}{Dt} \left[ ∫_{V_m(t)} B(\mathbf x,t)\ dV \right] = ∫_{V_m(t)} \left[ \frac{∂B}{∂t} + ∇∙(Bu) \right]dV $$ Left-Hand Side: $$ ∫_{CV} B\ dV = ∫_0 ^{4t} ∫_0 ^{2t} ∫_0 ^{t} (x\ +y\ +zt)\ dx\ dy\ dz = \frac12 x^2 |_0^t (8t^2)+\frac12 y^2 |_0^{2t}(4t^2)+(t) \frac12 z^2|_0^{4t} (2t^2)=4t^4+8t^4+16t^5=12t^4+16t^5 $$ Therefore; $$ \frac{D}{Dt} \left[ ∫_{CV} B\ dV \right]=48t^3+80t^4 $$ Right-Hand Side: $$∫_{V_m (t)} \frac{∂B}{∂t} dV=∫_0^{4t}∫_0^{2t}∫_0^t \frac∂{∂t} (x+y+zt)\ dx\ dy\ dz=∫_0^{4t} ∫_0^{2t}∫_0^t (z)\ dx\ dy\ dz=(t)(2t) \left( \frac12 z^2 |_0^{4t} \right)=16t^4 $$ Therefore; $$ ∫_{V_m (t)} \frac{∂B}{∂t} dV=16t^4 $$ By applying Divergence Theorem: $$ ∫_V div\ \mathbf F\ dV=∫_S \mathbf F\ d\mathbf S $$ $$ ∫_{V_m (t)} (\ ∇\ ∙\ (B\ \mathbf u)\ )\ dV = ∫_S (B\ \mathbf u)∙\ \hat{n}\ dS $$ In the direction of x-axis: $$ ∫_{S_x} (B\ {V_S})\ ∙\ \hat{n} \ dS = ∫_0^{4t}∫_0^{2t}(x+y+zt)|_{x=t} \left( \frac{dt}{dt} \right)\ dy\ dz\ = \left[ (t)(2t)(4t) + \left( \frac12 (2t)^2 \right) (4t)+(t)(2t) \left( \frac12 (4t)^2 \right) \right](1) = 8t^3 + 8t^3 + 16t^4 = 16t^3 + 16t^4 $$ In the direction of y-axis: $$ ∫_{S_y} (B\ {V_S})\ ∙\ \hat{n} \ dS = ∫_0^{4t}∫_0^t(x+y+zt)|_{y=2t} \left( \frac{d}{dt}(2t) \right) \ dx\ dz\ = \left[ \left( \frac12 (t)^2 \right) (4t) + (2t) (t) (4t) + (t) (t) \left( \frac12 (4t)^2 \right) \right] (2) = ( 2t^3 + 8t^3 + 8t^4 )(2) = 20t^3 + 16t^4 $$ In the direction of z-axis: $$ ∫_{S_z} (B\ {V_S})\ ∙\ \hat{n} ] dS = ∫_0^{2t}∫_0^t (x+y+zt)|_{z=4t}) \left( \frac{d}{dt}(4t) \right) \ dx\ dy\ = \left[ \left( \frac12 (t)^2 \right) (2t) +(t) \left( \frac12 (2t)^2 \right) + (t) (2t) (4t^2) \right] (4) = ( t^3 + 2t^3 + 8t^4 ) (4) = 12t^3 + 32t^4 $$ Therefore; $$ ∫_{V_m (t)} (\ ∇\ ∙\ (B\ \mathbf u)\ )\ dV = ( 16t^3 + 16t^4 ) + ( 20t^3 + 16t^4 ) + ( 12t^3 + 32t^4 ) = 48t^3 + 64t^4 $$ Right-Hand Side: $$ ∫_{V_m (t)} \frac{∂B}{∂t} dV + ∫_{V_m (t)} (\ ∇\ ∙\ (B\ \mathbf u)\ )\ dV = 16t^4 + (48t^3 + 64t^4) = 48t^3 + 80t^4 $$