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 determine the divergence of the vector field D, which is defined as:

[

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 the formula:

[

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

]

For our field:

– (P = 2xy)

– (Q = x^2)

– (R = 0)

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)

Now putting these together:

[

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