Gauss/Divergence Theorem

Question

Calculate ∫∫_S F • dS

F = ⟨ 3xy², xe^z , z³ ⟩

S is the surface of the solid bounded by the cylinder y² + z² = 16 and the planes x = 1 and x = 5

Answer 1

1. Find the Divergence of F

Given the vector field F= ⟨3xy², xe^z, z³⟩, the divergence is calculated as follows:

Compute the partial derivative of the x-component (3xy²) with respect to x:

This is 3y².

Compute the partial derivative of the y-component (xe^z) with respect to y:

This is 0, since xe^z does not depend on y.

Compute the partial derivative of the z-component (z³) with respect to z:

This is 3z².

So, the divergence of F is:

3y² + 3z².

2.Convert Divergence to Cylindrical Coordinates

Given divergence is 3y² + 3z².

In cylindrical coordinates:

y = r cos(θ)

z = r sin(θ)

y² = r² cos²(θ)

z² = r² sin²(θ)

Substitute y² and z²:

3y² = 3(r² cos²(θ))

3z² = 3(r² sin²(θ))

Add them: 3r² cos²(θ) + 3r² sin²(θ)

Factor out r²: 3r² (cos²(θ) + sin²(θ))

Since cos²(θ) + sin²(θ) = 1, it simplifies to 3r²

So, the divergence in cylindrical coordinates is 3r².

3. Set Up and Evaluate the Volume Integral

Volume Integral of 3r²:

The volume element in cylindrical coordinates is r , dr, dθ,dx

Set up the integral:

x ranges from 1 to 5

r ranges from 0 to 4

θ ranges from 0 to 2π

The integral is:

∫ (from x = 1 to 5) ∫ (from r = 0 to 4) ∫ (from θ = 0 to 2π) 3r² * r dθ dr dx

Simplify the Integrand:

3r² * r = 3r³

Perform the Integration:

1. Integrate with Respect to θ:

∫ (from θ = 0 to 2π) 3r³ dθ

=3r³ * (2π-0)

= 3r³ * 2π

= 6πr³

2. Integrate with Respect to r:

∫ (from r = 0 to 4) 6πr³ dr

Integral of r³ is r⁴ / 4

= 6π * [ (4⁴ / 4) - (0 / 4) ]

= 6π * 64

= 384π

3. Integrate with Respect to x:

∫ (from x = 1 to 5) 384π dx

= 384π * (5 - 1)

= 384π * 4

= 1536π

Thus, the surface integral is 1536π.

Answer 2

divF = 3y² + 3z²; I = ∫∫_SF•dS = ∫∫∫_v(3y² + 3z²)dxdydz; y = rcosθ; y = rsinθ; x = x; dxdydz = rdrdθddx; 0 ≤r ≤ 4

0 ≤ θ ≤ 2π. 1 ≤ x ≤ 5. I = ∫₁⁵dx ∫₀⁴3r³dr∫₀^2πdθ = x₁⁵3r⁴/4₀⁴θ₀^2π = (5 - 1)·3·64·2π = 1536π