∭▒〖∇.F ̅.dv〗=∭▒〖F ̅.ds〗
∇.F ̅ = ∂/∂x (3x+z^77 )+∂/∂y (y^2-sin x^(2 ) z)+∂/∂z (xz+〖ye〗^5x )
∇.F ̅ = =(3+z^77 )+(2y-sin x^(2 ) z)+(x+〖ye〗^5x )
∭▒〖∇.F ̅= 〗 ʃ^1 ʃ^3 ʃ^2 (3+z^77 )+(2y-sin x^(2 ) z)+(x+〖ye〗^5x ) .dx dy dz
= ʃ^1 ʃ^3 ʃ^2 x+5y . dx dy dz
ʃ^1 x.dx = [ x^2/2 ]^1 = [ 1/2 -0 ] = 1/2

Question

∭▒〖∇.F ̅.dv〗=∭▒〖F ̅.ds〗  	
∇.F ̅ = ∂/∂x (3x+z^77  )+∂/∂y (y^2-sin x^(2 ) z)+∂/∂z (xz+〖ye〗^5x  )	
∇.F ̅ = =(3+z^77  )+(2y-sin x^(2 ) z)+(x+〖ye〗^5x  )
∭▒〖∇.F ̅=  〗 ʃ^1 ʃ^3 ʃ^2 (3+z^77  )+(2y-sin x^(2 ) z)+(x+〖ye〗^5x  ) .dx dy dz
             = ʃ^1 ʃ^3 ʃ^2  x+5y . dx dy dz
    ʃ^1 x.dx = [  x^2/2  ]^1 = [  1/2  -0 ] =  1/2

fibonaccichick666 · Answer

you my friend need to learn LaTex lol gimme a min to decipher

anonymous · Answer

I have problem when I typed the question sorry sorry

fibonaccichick666 · Answer

lol it's cool

fibonaccichick666 · Answer

ok so what is the interval or C?

fibonaccichick666 · Answer

$$∭[∇.F̅.dv]=∭[F̅.ds]$$
$$∇.F̅=∂/∂x(3x+z77)+∂/∂y(y2−sinx(2)z)+∂/∂z(xz+〖ye〗5x)$$

$$∇.F̅==(3+z77)+(2y−sinx(2)z)+(x+〖ye〗5x)$$

$$∭〖∇.F̅=〗ʃ1ʃ3ʃ2(3+z77)+(2y−sinx(2)z)+(x+〖ye〗5x).dxdydz$$

            
$$=\int \int \int x+5y.dxdydz$$

   
$$ \int 1x.dx=[x2/2]1=[1/2−0]=1/2$$

anonymous · Answer

If D = (3x+z^77)ax + (y^2-sin x^2z)ay + (xz +ye^5x)az. Find the electric flux through the surface of the box;
0<x<1 ; 0<y<3 ; 0<z<2

fibonaccichick666 · Answer

so can you tell me what the normal vector is?

fibonaccichick666 · Answer

correction, unit normal vector sorry