Verify the divergence theorem for F= yi+xj+z

2k over the region

bounded by z=0,z=2,x
2 + y
2 = 4

This describes a cylindrical volume extending from z = 0 to z = 2 with a circular cross-section of radius 2.

Step 2: Calculate the Divergence of F

The divergence of a vector field **F = P i + Q j + R k** is given by:

∇ · F = ∂P/∂x + ∂Q/∂y + ∂R/∂z

For our vector field:

• P = y
• Q = x
• R = z²

Now, we compute the partial derivatives:

• ∂P/∂x = 0
• ∂Q/∂y = 1
• ∂R/∂z = 2z

Thus, the divergence is:

∇ · F = 0 + 1 + 2z = 1 + 2z

Step 3: Volume Integral of the Divergence

Next, we need to compute the volume integral of the divergence over the cylindrical region:

∬_V (∇ · F) dV = ∬_V (1 + 2z) dV

In cylindrical coordinates, where x = r cos(θ), y = r sin(θ), and z = z, the volume element is dV = r dr dθ dz. The limits for r will be from 0 to 2, for θ from 0 to 2π, and for z from 0 to 2.

Setting Up the Integral

The volume integral becomes:

∫_0^2 ∫_0^{2π} ∫_0^2 (1 + 2z) r dr dθ dz

Calculating the Integral

We can break this integral into two parts:

∫_0^2 ∫_0^{2π} ∫_0^2 r dr dθ dz + 2 ∫_0^2 ∫_0^{2π} ∫_0^2 z r dr dθ dz

Calculating the first part:

• ∫_0^2 r dr = [r²/2]_0^2 = 2
• ∫_0^{2π} dθ = 2π

So, the first integral evaluates to:

2 * 2π * 2 = 8π

Now for the second part:

• ∫_0^2 z dz = [z²/2]_0^2 = 2
• ∫_0^{2π} dθ = 2π

Thus, the second integral evaluates to:

2 * 2π * 2 = 8π

Combining both parts gives:

8π + 8π = 16π

Step 4: Surface Integral of F

Now, we need to compute the surface integral of **F** over the closed surface bounding the volume. This surface consists of three parts: the top (z = 2), the bottom (z = 0), and the curved side of the cylinder.

Top Surface (z = 2)

On the top surface, **F = yi + xj + 4k**. The outward normal vector is **k**. The flux through the top surface is:

∬_S F · n dS = ∬_S (4) dS = 4 * Area of the circle = 4 * π(2²) = 16π

Bottom Surface (z = 0)

On the bottom surface, **F = yi + xj + 0k**. The outward normal vector is **-k**. The flux through the bottom surface is:

∬_S F · n dS = ∬_S (0) dS = 0

Curved Surface of the Cylinder

On the curved surface, we parameterize it using cylindrical coordinates. The outward normal vector is **r**. The flux through the curved surface can be computed, but due to symmetry, it will also yield a contribution of **0**.

Final Calculation

Adding the contributions from all surfaces:

Flux through top + Flux through bottom + Flux through curved surface = 16π + 0 + 0 = 16π

Verification of the Divergence Theorem

We found that the volume integral of the divergence is **16π**, and the surface integral of **F** over the closed surface is also **16π**. Since both sides are equal, we have verified the divergence theorem for the given vector field and region.