Question

Suppose f⃗ is a vector field with div(f⃗ (x,y,z))=−3. use the divergence theorem to calculate the flux of the vector field f⃗ out of the closed, outward-oriented cylindrical surface s of height 7 and radius 5 that is centered about the z-axis with its base in the xy-plane.

Answer 1

To calculate the flux using the divergence theorem, multiply the constant divergence of the vector field (-3) with the volume of the cylinder (175π). The resulting flux through the cylindrical surface is -525π.

Step-by-step explanation:

The question involves using the divergence theorem to calculate the flux of a vector field out of a cylindrical surface. The divergence of the vector field f is given as -3. The divergence theorem, also known as Gauss's theorem, relates the flux through a closed surface to the divergence of a vector field within the volume enclosed by the surface. According to the theorem:

Φ = ∫_S f · dA = ∫_V div(f) dV

Given the height of the cylinder is 7 and the radius is 5, we can find the volume V of the cylinder using the formula V = πr^2h. Substituting the height and radius, we get V = π(5)^2(7) = 175π.

Since the divergence is constant and equal to -3, the integral over the volume simplifies to a multiplication by the volume:

Flux (Φ) = div(f) × Volume V = -3 × 175π = -525π

This result represents the total flux of the vector field out of the cylindrical surface.

Answer 2

The surface
`S` is the boundary of a cylindrical region
`R` that can be expressed in cylindrical coordinates by the set of points

`\{(r,\theta,z)\mid0\le r\le5\,\land\,0\le\theta\le2\pi\,\land\,0\le z\le7\}`

By the divergence theorem, the flux of
`\vec f` across
`S` is given by the integral of
`\mathrm{div}(\vec f)=-3` over
`R`:

`\displaystyle\iiint_R-3\,\mathrm dV=-3\int_(z=0)^(z=7)\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=5)r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz`

`=-21\pi r^2\bigg|_(r=0)^(r=5)=-525\pi`