Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.

To apply the Divergence Theorem, we first need to compute the divergence of the vector field ( mathbf{D} = 2xy mathbf{i} + x^2 mathbf{j} ).

The divergence of a vector field ( mathbf{D} = P mathbf{i} + Q mathbf{j} + R mathbf{k} ) is given by:

[

nabla cdot mathbf{D} = frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z}

]

For our vector field ( mathbf{D} ):

– ( P = 2xy )

– ( Q = x^2 )

– ( R = 0 )

Now we compute the partial derivatives:

1. ( frac{partial P}{partial x} = frac{partial (2xy)}{partial x} = 2y )
2. ( frac{partial Q}{partial y} = frac{partial (x^2)}{partial y} = 0 )
3. ( frac{partial R}{partial z} = frac{partial (0)}{partial z} = 0 )

Thus, the divergence is:

[

nabla cdot mathbf{D} = 2y + 0 + 0 = 2y