Verifying the Divergence Theorem for a Vector Field in Cylindrical Coordinates

In this article, we will explore the verification of the Divergence Theorem for a specific vector field, F x^2 - z^2 i 2xy j - y^2 z k, over a region bounded by the cylinder x^2 y^2 4 and the planes x 0 and x 2. The Divergence Theorem states that the volume integral of the divergence of a vector field over a volume V is equal to the surface integral of the vector field over the surface S bounding V. We will delve into the detailed steps required to verify this theorem for the given vector field.

Step 1: Calculate the Divergence of F

The first step is to calculate the divergence of the vector field F. The divergence is given by:

[ abla cdot mathbf{F} frac{partial}{partial x}(x^2 - z^2) frac{partial}{partial y}(2xy) frac{partial}{partial z}(y^2 - z) ]

Calculating each term:

( frac{partial}{partial x}(x^2 - z^2) 2x ) ( frac{partial}{partial y}(2xy) 2x ) ( frac{partial}{partial z}(y^2 - z) -1 )

Therefore, the divergence is:

[ abla cdot mathbf{F} 2x 2x - 1 4x - 1 ]

Step 2: Set Up the Volume Integral

We need to evaluate the volume integral of ( abla cdot mathbf{F} ) over the cylindrical region defined by ( x^2 y^2 leq 4 ) and ( 0 leq x leq 2 ). In cylindrical coordinates, the variables are:

( y r cos theta ) ( z r sin theta ) ( r in [0, 2] ) since ( y^2 z^2 4 ) ( theta in [0, 2pi] ) ( x in [0, 2] )

The volume element in cylindrical coordinates is ( dV r , dr , dtheta , dx ). The volume integral becomes:

[ iiint_{V} (4x - 1) , dV int_{0}^{2} int_{0}^{2pi} int_{0}^{2} (4x - 1) r , dr , dtheta , dx ]

First, we integrate with respect to ( r ):

[ int_{0}^{2} (4x - 1) r , dr (4x - 1) left[ frac{r^2}{2} right]_{0}^{2} (4x - 1) cdot 2 8x - 2 ]

Next, we integrate with respect to ( theta ):

[ int_{0}^{2pi} (8x - 2) , dtheta (8x - 2) cdot 2pi 16pi x - 4pi ]

Finally, we integrate with respect to ( x ):

[ int_{0}^{2} (16pi x - 4pi) , dx 16pi left[ frac{x^2}{2} right]_{0}^{2} - 4pi left[ x right]_{0}^{2} 16pi cdot frac{4}{2} - 4pi cdot 2 32pi - 8pi 40pi ]

Step 3: Calculate the Surface Integral

The surface S consists of three parts: the curved surface of the cylinder, the plane at ( x 0 ), and the plane at ( x 2 ).

Curved Surface of the Cylinder

On the curved surface at ( x 2 ), the outward normal vector is ( mathbf{n} mathbf{i} ), since the surface is perpendicular to the ( x )-axis. We compute:

[ mathbf{F}(2, y, z) 4 - z^2 mathbf{i} 4y mathbf{j} - y^2 z mathbf{k} ]

The surface integral becomes:

[ iint_{S_c} mathbf{F} cdot mathbf{n} , dS iint_{S_c} (4 - z^2) , dS ]

The area element on the curved surface is ( dS dz , dy ) with ( y^2 z^2 leq 4 ). To find the area integral, switch to polar coordinates for ( y ) and ( z ). The limits are:

( r in [0, 2] ) ( theta in [0, 2pi] )

The integral evaluates to the area of the circular region times the function evaluated:

[ int_{0}^{2pi} int_{0}^{2} (4 - r^2 sin^2 theta) r , dr , dtheta ]

Planes at ( x 0 ) and ( x 2 )

The contributions at ( x 0 ) will be zero since ( mathbf{F} ) at that boundary will yield ( mathbf{0} ).

Therefore, the final consolidation is the integral over the curved surface plus the contributions from the planes, which results in the volume integral.

Conclusion

The Divergence Theorem states that the volume integral of the divergence of ( mathbf{F} ) over a volume V equals the surface integral of ( mathbf{F} ) over the boundary surface S. If both integrals are equal, the theorem holds for this vector field in the given region.