To find the value of the divergence of vector field ( mathbf{A} = xy^2 hat{i} + y^3 hat{j} + y^2z hat{k} ) over the cuboid defined by the intervals (0 < x < 1), (0 < y < 1), and (0 < z < 1) using the divergence theorem, we first need to compute the divergence of ( mathbf{A} ).

The divergence of a vector field ( mathbf{A} = Phat{i} + Qhat{j} + Rhat{k} ) is given by:

[ nabla cdot mathbf{A} = frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z} ]

For ( mathbf{A} = xy^2 hat{i} + y^3 hat{j} + y^2z hat{k} ),

[ P = xy^2, Q = y^3, R = y^2z ]

Computing the partial derivatives,

[ frac{partial P}{partial x} = y^2 ]

[ frac{partial Q}{partial y} = 3y^2 ]

[ frac{partial R}{partial z} = y^2

To compute Gauss’s law using the given electric flux density, (D = frac{10rho^3}{4} hat{i}) in cylindrical coordinates, we follow the integral form of Gauss’s law for electric fields, which states:

[

oint_S mathbf{D} cdot dmathbf{A} = Q_{text{enc}}

]

Where:

– (oint_S mathbf{D} cdot dmathbf{A}) is the electric flux through a closed surface (S),

– (Q_{text{enc}}) is the total electric charge enclosed by the surface.

Given that (D) is in the direction of (hat{i}), which implies it’s purely radial in cylindrical coordinates, this simplifies the problem since we only need to consider the component of (mathbf{D}) that is normal to the surface (S). For a cylindrical surface with radius (rho) and height from (z=0) to (z=5), the relevant surface area for the flux calculation is the curved surface area, because the electric flux density is radially outward and will not penetrate the top and bottom surfaces normally.

The curved surface area of a cylinder is (A = 2pirho h), where (rho) is the radius of the cylinder and (h) is the height. For (rho = 4

A field in which the net electric flux through any closed surface is zero is characteristic of an electrostatic field in equilibrium. Specifically, this principle is encapsulated by Gauss’s Law in electrostatics, which states that the total electric flux through a closed surface is proportional to the charge enclosed by the surface. Therefore, if a test charge moves around any closed path in such a field and the net work done is zero, it implies that the electric field is conservative. However, the precise terminology you’re looking for, which describes the scenario where the net work done on a test charge over a closed path is zero, refers to a conservative electric field. Yet, your question hints at the concept encapsulated by Gauss’s Law for electrostatics, where the net electric flux out of any closed surface is proportional to the charge enclosed by the surface. If there’s no charge within the closed surface or if the positive and negative charges inside cancel out, the net electric flux would be zero, indicating an electrostatic equilibrium condition within that closed surface.

The question involves converting a vector given in spherical coordinates to Cartesian coordinates and then evaluating it at a specific Cartesian point. However, there seems to be a misunderstanding in how the question is posed. Let’s clarify the concepts involved before directly addressing the problem as it’s presented.

1. Clarification on Coordinates and Vector Fields:

– Spherical coordinates are typically denoted by ((rho, theta, phi)), where (rho) is the radial distance, (theta) is the azimuthal angle, and (phi) is the polar angle.

– Cartesian coordinates are denoted by ((x, y, z)).

– The vector field given, (mathbf{B} = left(frac{10}{r}mathbf{i} + (r costheta) mathbf{j} + mathbf{k} right)), appears to be defined with an assumption of some transformations that aren’t standard for spherical to Cartesian conversion. Specifically, (r) in the context given seems to represent a position vector magnitude, but without clearer definition, its conversion context is ambiguous. In standard spherical coordinates, (r = rho), the radial distance from the origin.

2. Conversion Process:

To approach the instruction literally, converting a vector from spherical to Cartesian coordinates generally involves using transformation equations based on the definitions of the spherical coordinates:

– (x = rho sinphi

Vector transformation followed by coordinate point substitution, and vice versa, giving the same result, indicates the consistency and interchangeability of applying linear transformations in algebra. This process can be visualized within the context of linear algebra, particularly when dealing with transformations in spaces like ( mathbb{R}^n ).

Given a vector (mathbf{v}) in ( mathbb{R}^n ) and a linear transformation ( T: mathbb{R}^n rightarrow mathbb{R}^m ), applying ( T ) to (mathbf{v}) and then substituting coordinates (or vice versa) leads to the same result due to the properties of linear transformations. This involves the transformation matrix ( A ) associated with ( T ), which acts on (mathbf{v}) to produce a new vector in ( mathbb{R}^m ).

The process is as follows:

1. Vector Transformation: Apply the transformation ( T ) to vector (mathbf{v}), resulting in ( T(mathbf{v}) = Amathbf{v} ), where ( A ) is the transformation matrix.

2. Coordinate Point Substitution: After applying the transformation, we can substitute the coordinates of (mathbf{v}) into the resulting vector to find its new location in ( mathbb{R}^m ).

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