Question

In Exercises \(5-8\), a closed surface \(S\) enclosing a domain \(D\) and a vector field \(\vec{F}\) are given. Verify the Divergence Theorem on \(\mathcal{S} ;\) that is, show \(\iint_{S} \vec{F} \cdot \vec{n} d S=\iiint_{D} \operatorname{div} \vec{F} d V\). \(\mathcal{S}\) is the surface bounding the domain \(D\) enclosed by the plane \(z=2-x / 2-2 y / 3\) and the coordinate planes in the first octant; \(\vec{F}=\left\langle x^{2}, y^{2}, x\right\rangle\).

Short answer

The Divergence Theorem holds for this domain.

Step-by-step solution

Parameterize the Surface and Calculate the Bounds

First, determine the intersection of the plane with the coordinate axes to identify the limits for integration. For the given plane equation, set each coordinate to zero to find the intercepts:- Let \(z = 0\) to find the line in the \(xy\)-plane: \(0 = 2 - \frac{x}{2} - \frac{2y}{3}\). Solving this gives \(x = 4\) and \(y = 3\).From the intercepts, the domain \(D\) is a triangular region in the \(xy\)-plane with vertices at \((0, 0), (4, 0), (0, 3)\). The upper bound for \(z\) is given by the plane equation, \(z = 2 - \frac{x}{2} - \frac{2y}{3}\).

Compute Divergence of \\\(\

The divergence of a vector field \(\vec{F}(x, y, z) = \langle x^2, y^2, x \rangle\) is computed as:\[\text{div } \vec{F} = \frac{\partial}{\partial x} (x^2) + \frac{\partial}{\partial y} (y^2) + \frac{\partial}{\partial z} (x)\]Calculating each partial derivative gives:\[\text{div } \vec{F} = 2x + 2y + 0 = 2x + 2y\]

Evaluate the Volume Integral

Integrate the divergence over the domain \(D\) using the bounds determined:\[\iiint_{D} \operatorname{div} \vec{F} \, dV = \int_{0}^{3} \int_{0}^{4-\frac{3}{2}y} \int_{0}^{2 - \frac{x}{2} - \frac{2y}{3}} (2x + 2y) \, dz \, dx \, dy\]Perform the integration step by step:1. Integrate with respect to \(z\): The limits for \(z\) are from \(0\) to \(2 - \frac{x}{2} - \frac{2y}{3}\), so the integral becomes: \[ \int_{0}^{2 - \frac{x}{2} - \frac{2y}{3}} (2x + 2y) \, dz = (2x + 2y)(2 - \frac{x}{2} - \frac{2y}{3}) \]2. Integrate the result with respect to \(x\), from \(x = 0\) to \(x = 4 - \frac{3}{2}y\).3. Final integration with respect to \(y\), from \(y = 0\) to \(y = 3\).

Calculate the Surface Integral

The surface integral requires evaluating \(\iint_{S} \vec{F} \cdot \vec{n} \, dS\), where \(\vec{n}\) is the outward normal to the surface \(S\):Since the surface consists of the plane and the three coordinate planes, split \(S\) into these parts:1. **Top plane**: The plane \(z = 2 - \frac{x}{2} - \frac{2y}{3}\). Use the normal obtained from the gradient as \(\vec{n} = \langle -\frac{1}{2}, -\frac{2}{3}, 1 \rangle\).2. **Coordinate planes**: Evaluate where \(x = 0\), \(y = 0\), and \(z = 0\). These have normals \(\langle 1, 0, 0 \rangle\), \(\langle 0, 1, 0 \rangle\), and \(\langle 0, 0, 1 \rangle\) respectively. For planes of form \(x=4-\frac{3}{2}y\) and \(y=3\), apply necessary transformations. Combine the results for all contributing surfaces.

Verify The Divergence Theorem

After evaluating both sides of the Divergence Theorem:1. Sum up the results from the surface integral computation and the volume integral computation.2. Verify that both integrals are equal, confirming the Divergence Theorem for \(S\).This completes the verification, showing that the divergence theorem holds true for the region. Each calculated integral and the sum must match to conclude that the theorem is correct for this example.

Key concepts

vector fields

A vector field is like a map that contains a vector (an arrow with direction and magnitude) for every point in a space. Think about a weather map with arrows showing wind direction and speed. In mathematics and physics, vector fields are useful tools for describing physical phenomena like gravitational, electric, and magnetic forces. In our specific problem, we are dealing with the vector field \(\vec{F} = \langle x^2, y^2, x \rangle\). Each component of the vector assigns a force at each point in space.

• \(x^2\): The first component corresponds to the force in the x-direction.
• \(y^2\): The second component deals with the force in the y-direction.
• \(x\): The third component indicates the force in the z-direction.

Understanding the behavior of this vector field helps us set up integrals that we will later evaluate, especially when applying the Divergence Theorem, which links these vector fields to the surface and volume integrals.

surface integrals

Surface integrals are a way to integrate over a surface, as opposed to along a line or through a volume. They are very useful when we want to calculate things like the total flux of a vector field across a surface. Imagine water flowing across a fishing net; we would want to know the total amount that passes through the net.To evaluate a surface integral, we need to know the normal vector \(\vec{n}\), which tells us the direction perpendicular to the surface at each point. This surface integral is expressed as \(\iint_{S} \vec{F} \cdot \vec{n} \, dS\), where \(S\) is our surface.Our closed surface \(S\) includes a plane and three coordinate planes:

• Top Plane: Calculating the component of the vector field \(\vec{F}\) crossing the boundary defined by our given plane.
• Coordinate Planes: We must also consider where \(x = 0\), \(y = 0\), and \(z = 0\), each having their normals pointing outward.

These computations help us find the net amount of the vector field through the surface, crucial for verifying the Divergence Theorem.

volume integrals

Volume integrals extend the concept of integration to a three-dimensional space, allowing us to integrate over a volume rather than just area or line. These integrals are essential when using the Divergence Theorem to connect the surface integral to the behavior of the vector field inside a volume.The volume integral we are considering takes the divergence of our vector field, \(\operatorname{div} \vec{F}\), over the domain \(D\) enclosed by surface \(S\). This is given by \(\iiint_{D} \operatorname{div} \vec{F} \, dV\).To perform this integral, you first:

• Determine the bounds for \(x\), \(y\), and \(z\) based on the geometry of the region.
• Compute the divergence \(\operatorname{div} \vec{F} = 2x + 2y\) which sums up how much the vector field spreads out from a point.
• Carefully perform the integration inside these bounds, summing the results across the volume.

This volume integral gives us a measure of the total flux emerging from a volume through the vector field inside it, ensuring our understanding ties together with the theorem.

closed surfaces

Closed surfaces are surfaces that completely enclose a space, like the shell of a ball or a cylinder with both ends capped. These surfaces don't have any edges or openings and make up the boundaries for a three-dimensional region where we can apply the Divergence Theorem.In our exercise, we have a specific closed surface \(\mathcal{S}\), which encloses the domain \(D\). It is formed by:

• A given plane: Extends over a region, and its equation helps define the top part of our surface.
• Coordinate planes: The \(x\)-, \(y\)-, and \(z\)-axes complete the closed boundary in the first octant.

Closed surfaces are essential for applications like the Divergence Theorem, as they ensure that every divergence computed inside a region has a corresponding flux across the region's boundary. This closure is key to establishing that the flow through the volume's surface equals the flow coming from within it, proving that both sides of the equation in the theorem will balance out.