Question

Use the Divergence Theorem to compute the net outward flux of the following fields across the given surfaces \(S\). $$\begin{aligned} &\mathbf{F}=\left\langle y-2 x, x^{3}-y, y^{2}-z\right\rangle ; S \text { is the sphere }\\\ &\left\\{(x, y, z): x^{2}+y^{2}+z^{2}=4\right\\} \end{aligned}$$

Short answer

Answer: The net outward flux of the vector field \(\mathbf{F}\) through the spherical surface \(S\) is \(-\frac{64}{3}\pi\).

Step-by-step solution

Find the divergence of the vector field \(\mathbf{F}\)

To find the divergence of \(\mathbf{F}\), we need to compute its partial derivatives with respect to \(x, y, z\) and add them up. We have: $$\nabla \cdot \mathbf{F} = \frac{\partial (y-2x)}{\partial x} + \frac{\partial (x^3-y)}{\partial y} + \frac{\partial (y^2-z)}{\partial z} = -2 + 1 - 1 = -2$$

Apply the Divergence Theorem

Now that we have the divergence of the vector field, we can apply the Divergence Theorem to find the outward flux through the surface \(S\). The theorem states: $$\iint_{S} \mathbf{F} \cdot \mathbf{n} dS = \int_{V} \nabla \cdot \mathbf{F} dV$$ where \(\mathbf{n}\) is the outward unit normal vector to the surface, and \(V\) is the volume enclosed by the sphere. In this case, it's a sphere of radius 2 centred at the origin. The volume of the sphere is given by the formula \(V_{sphere} = \frac{4}{3}\pi r^3\). Plugging in the radius of 2, we get: $$V = \frac{4}{3}\pi (2)^3 = \frac{32}{3}\pi$$ Now, we can plug in the divergence of the vector field \(\nabla \cdot \mathbf{F} = -2\): $$\int_{V} \nabla \cdot \mathbf{F} dV = -2 \int_{V} dV = -2 V_{sphere} = -2 \frac{32}{3}\pi = -\frac{64}{3}\pi$$ Therefore, the net outward flux of the vector field \(\mathbf{F}\) through the spherical surface \(S\) is equal to \(-\frac{64}{3}\pi\).

Key concepts

Vector Field Flux

Vector field flux is fundamental in understanding various physical concepts like electromagnetism and fluid dynamics. When we consider a vector field, such as \( \mathbf{F}=\langle y-2x, x^3-y, y^2-z \rangle \), it represents a quantity with both magnitude and direction at every point in space. Imagine that this vector field represents the flow of a fluid or the strength and direction of a magnetic field.

The key question we answer with flux is: how much of this 'stuff' (fluid, air, magnetic field, etc.) is passing through a certain area? To find the flux through a surface, such as a sphere with radius 2, we 'count' the number of 'field lines' that pass through this surface. Mathematically, this is done by taking the dot product of the field \(\mathbf{F}\) with the outward unit normal vector \(\mathbf{n}\) of the surface and integrating over the entire surface \(S\).

However, calculating this directly can be laborious, which is why the Divergence Theorem is so powerful. It lets us convert this surface integral into a typically easier volume integral over the region enclosed by the surface.

Partial Derivatives

Partial derivatives are indispensable when dealing with multivariable functions, such as vector fields. They provide the rate at which a function changes with respect to one of its variables while holding the others constant. In the three-dimensional vector field \(\mathbf{F}\), we have three components, each of which is a function of \(x, y,\) and \(z\). To find the divergence of \(\mathbf{F}\), which is needed for the Divergence Theorem, we compute the partial derivatives of each component with respect to its corresponding variable.

Illustratively, for our vector field \(\mathbf{F}\), the divergence \(abla \cdot \mathbf{F}\) is calculated as follows: \[-2 + 1 - 1 = -2\]. This scalar quantity gives us a measure of how much the vector field 'sources' or 'sinks' at each point in space.

Outward Unit Normal Vector

The outward unit normal vector, denoted as \(\mathbf{n}\), is a vector that is perpendicular to the surface at a given point and points away from the enclosed volume. In the case of a sphere, the outward unit normal can be obtained by normalizing the vector that goes from the center of the sphere to a point on the surface. Since the sphere is centered at the origin in our exercise, this vector coincides with the position vector of the point.

If we were to calculate the flux directly, we would need to use this \(\mathbf{n}\) in the dot product with \(\mathbf{F}\). Instead, applying the Divergence Theorem simplifies our job, meaning we do not need to explicitly calculate this normal vector for each point on the surface.

Volume of the Sphere

The volume of a sphere is an expression of how much three-dimensional space the sphere occupies and is vital for applying the Divergence Theorem. For a sphere with radius \(r\), the volume can be calculated using the formula \(V_{sphere} = \frac{4}{3}\pi r^3\).

Knowing the volume is essential because according to the Divergence Theorem, we integrate the divergence of \(\mathbf{F}\) over the volume enclosed by surface \(S\), rather than integrating \(\mathbf{F}\cdot \mathbf{n}\) over the surface itself. For instance, the sphere in the exercise problem has a radius of 2, giving us a volume of \(\frac{32}{3}\pi\). This simplifies calculations greatly, reducing the problem to multiplying the divergence by the volume of the sphere to find the net outward flux.