Question

Use the Divergence Theorem to compute the net outward flux of the following fields across the given surfaces \(S\). \(\mathbf{F}=\langle x, y, z\rangle ; S\) is the surface of the cone \(z^{2}=x^{2}+y^{2},\) for \(0 \leq z \leq 4,\) plus its top surface in the plane \(z=4\)

Short answer

Answer: The net outward flux of the vector field through the surface S is \(192\pi\).

Step-by-step solution

Find the Divergence

Compute the divergence of the vector field \(\mathbf{F} = \langle x, y, z \rangle\). The divergence is given by: \(div(\mathbf{F}) = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3.\)

Triple Integral Bounds

Define the bounds for the triple integral using cylindrical coordinates \((r, \theta, z)\). The cone can be described as: \(z=r,\) where \(0\leq r \leq 4\) and \(0\leq z\leq 4\). The surface \(S\) consists of the lateral surface of the cone and the top disk, so we have the following limits of integration: \(r\) ranging from \(0\) to \(4\), \(\theta\) ranging from \(0\) to \(2\pi\), \(z\) ranging from \(0\) to \(4\).

Convert to Cylindrical Coordinates

Convert the divergence and volume element into cylindrical coordinates. The divergence is already a scalar value, so conversion is not needed. The volume element in cylindrical coordinates is given by: \(dV = r \, dr \, d\theta \, dz\).

Evaluate the Triple Integral

Compute the net outward flux using the Divergence Theorem, which states: \(\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} div(\mathbf{F}) \, dV.\) Evaluating the triple integral with the divergence value and volume element gives: \(\iiint_{V} 3 \, r \, dr \, d\theta \, dz = 3 \int_{0}^{2\pi} \int_{0}^{4} \int_{0}^{4} r \, dr \, d\theta \, dz.\) Evaluating each integral separately: \(\int_{0}^{2\pi} d\theta = 2\pi\) \(\int_{0}^{4} dz = 4\) \(\int_{0}^{4} r \, dr = \frac{1}{2}r^2 \Big|_{0}^{4} = 8\).

Compute the Net Outward Flux

Multiply the result of each integral to find the net outward flux. Net outward flux \(= 3(2\pi)(4)(8) = 192\pi\). Hence, the net outward flux of the vector field \(\mathbf{F} = \langle x, y, z \rangle\) through the given surface S is \(192\pi\).

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a popular system in mathematics, especially useful for dealing with problems involving symmetric shapes like cylinders and cones. They extend the concept of polar coordinates into three dimensions, making them ideal for situations where standard Cartesian coordinates might be less intuitive.

Cylindrical coordinates consist of three values:

• \(r\): the radial distance from the origin to the projection of the point onto the \(xy\)-plane.
• \(\theta\): the angle between the positive \(x\)-axis and the line connecting the origin to the projection of the point onto the \(xy\)-plane.
• \(z\): the height of the point above the \(xy\)-plane, the same as in Cartesian coordinates.

This is why they are suitable for problems involving the symmetry around an axis, like the cone in our exercise. By using cylindrical coordinates, we can easily describe the cone with the equation \(z = r\), indicating that the height \(z\) increases at the same rate as \(r\) extends outward in the \(xy\)-plane.

Net Outward Flux

The concept of net outward flux is crucial in vector calculus and relates directly to the Divergence Theorem. It measures the total quantity of a vector field that passes out of a closed surface. Think of it like measuring how much fluid is flowing out of a surface area.

In practical terms, the net outward flux is what you calculate using the surface integral of the vector field across a surface, essentially summing up how the field behaves over every tiny piece of that surface. When the Divergence Theorem is applicable, the computation becomes simpler by transforming the surface integral into a volume integral. The theorem states that the net outward flux through a closed surface \(S\) is equivalent to the integral of the divergence of the field inside the volume \(V\) enclosed by \(S\). This is captured by the formula:
\[\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV\]The exercise involves finding this flux for the vector field \(\mathbf{F} = \langle x, y, z \rangle\) across the surface of a cone and its top disk by using the Divergence Theorem, transforming the problem into calculating a triple integral.

Triple Integral

A triple integral extends the concept of an integral to functions of three variables, allowing one to compute volumes and related quantities in three-dimensional space.

In our exercise, we use a triple integral to determine the net outward flux from the volume of the cone. Using cylindrical coordinates simplifies the integration since the boundary conditions of the cone fit this coordinate system naturally.

To conduct this integration, boundaries must be defined for each coordinate, much like using limits in single integrals:

• \(r\) ranges from 0 to 4: It moves radially outward from the cone's tip to its base.
• \(\theta\) ranges from 0 to \(2\pi\): It sweeps full rotations around the \(z\)-axis.
• \(z\) ranges from 0 to 4: Vertically from the cone's tip to its top.

When integrating, the volume element \(dV = r \, dr \, d\theta \, dz\) is used, considering the extra \(r\) factor that accounts for the radial measure in cylindrical coordinates. The triple integral's setup is crucial for calculating the flux by ensuring the integration covers the entire volume involved. By solving, we simplify to the product of results from each integral:
\(3 \cdot \left(2\pi\right) \cdot 4 \cdot 8 = 192\pi\), deducing the net outward flux.