Problem 384 - Calculus Volume 3

Question

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral \(\int_{S} \mathbf{F} \cdot \mathbf{n} d s\) for the given choice of \(\mathbf{F}\) and the boundary surface \(S\). For each closed surface, assume \(\mathbf{N}\) is the outward unit normal vector. [T] Surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S},\) where \(S\) is the solid bounded by paraboloid \(z=x^{2}+y^{2}\) and plane \(z=4\) and \(\mathbf{F}(x, y, z)=\left(x+y^{2} z^{2}\right) \mathbf{i}+\left(y+z^{2} x^{2}\right) \mathbf{j}+\left(z+x^{2} y^{2}\right) \mathbf{k}\)

Step-by-Step Solution

Answer

The surface integral is \(24\pi\).

Step 1: Understanding the Problem

Our task is to compute the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\) over the surface \(S\), which is given by the paraboloid \(z=x^2+y^2\) bounded by the plane \(z=4\). The vector field is \(\mathbf{F}(x, y, z) = (x + y^2 z^2) \mathbf{i} + (y + z^2 x^2) \mathbf{j} + (z + x^2 y^2) \mathbf{k}\). We will use the Divergence Theorem to solve this, which converts a surface integral into a volume integral.

Step 2: Divergence Theorem Application

The Divergence Theorem states that \(\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV\), where \(abla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\). Here, \(V\) is the volume enclosed by surface \(S\). We need to compute \(abla \cdot \mathbf{F}\) for the given vector field.

Step 3: Calculating the Divergence

The divergence \(abla \cdot \mathbf{F}\) of the vector field \(\mathbf{F}\) is computed as follows: \[ abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x + y^2 z^2) + \frac{\partial}{\partial y}(y + z^2 x^2) + \frac{\partial}{\partial z}(z + x^2 y^2) \]Calculating each term gives:\[ \frac{\partial}{\partial x}(x + y^2 z^2) = 1 \]\[ \frac{\partial}{\partial y}(y + z^2 x^2) = 1 \]\[ \frac{\partial}{\partial z}(z + x^2 y^2) = 1 \]Therefore, \(abla \cdot \mathbf{F} = 1 + 1 + 1 = 3\).

Step 4: Setting Up the Volume Integral

Now we will set up the volume integral \(\iiint_{V} 3 \, dV\) over the region \(V\) bounded by the paraboloid and the plane. In cylindrical coordinates, the paraboloid is described by \(z = r^2\) and the plane by \(z = 4\). Thus, \[ 0 \leq r \leq 2 \] \[ 0 \leq \theta \leq 2\pi \]\[ r^2 \leq z \leq 4 \]

Step 5: Evaluating the Volume Integral

The volume integral becomes \[ \iiint_{V} 3 \, dV = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} 3 \, dz \, r \, dr \, d\theta \]. Evaluate the integral step-by-step:Integrate with respect to \(z\): \[ \int_{r^2}^{4} 3 \, dz = 3[z]_{r^2}^{4} = 3(4 - r^2) \]Plug into the next integrals: \[ \int_{0}^{2} 3(4 - r^2)r \, dr = 12 \int_{0}^{2} r \, dr - 3 \int_{0}^{2} r^3 \, dr \]Evaluate the simpler integrals:For \(\int_{0}^{2} r \, dr = [\frac{r^2}{2}]_{0}^{2} = 2\)For \(\int_{0}^{2} r^3 \, dr = [\frac{r^4}{4}]_{0}^{2} = 4\)Combine results: \[ 12 \cdot 2 - 3 \cdot 4 = 24 - 12 = 12 \]Finally, integrate over \(\theta\): \[ \int_{0}^{2\pi} 12 \, d\theta = 12\theta \bigg|_{0}^{2\pi} = 12 \cdot 2\pi = 24\pi \].

Step 6: Conclusion of Solution

Applying the Divergence Theorem, the value of the surface integral \(\iint_{S} \mathbf{F} \cdot d\mathbf{S}\) is equal to the evaluated volume integral result, which is \(24\pi\).

Key Concepts

Surface Integral

A surface integral is an extension of multiple integrals; it allows us to integrate over a surface instead of a curve or a region in the plane. In the context of vector calculus, a surface integral like \(\iint_{S} \mathbf{F} \cdot d\mathbf{S}\) calculates the flow of a vector field \(\mathbf{F}\) across a surface \(S\). Here, \(d\mathbf{S}\) represents an infinitesimal piece of the surface with a direction given by the surface's normal vector.
In physical terms, surface integrals can represent the total flux across a boundary, such as how much of a fluid passes through a surface. This concept is closely related to the Divergence Theorem, which simplifies the calculation by converting it to a volume integral over the space enclosed by the surface.

Vector Field

A vector field is a construction in mathematics that assigns a vector to every point in a subset of space. This vector could represent a force like gravity, a velocity of fluid flow, or any quantity that has direction and magnitude at any point. The vector field given in our problem is \(\mathbf{F}(x, y, z) = (x + y^2 z^2) \mathbf{i} + (y + z^2 x^2) \mathbf{j} + (z + x^2 y^2) \mathbf{k}\).
Understanding the components:

• Each component of the vector field involves some relationship between \(x\), \(y\), and \(z\).
• This vector field changes in space and such representations are critical in modeling physical phenomena.

To solve problems involving vector fields, calculus operations like the gradient, divergence, and curl are commonly used. In our context, computing the divergence helps convert complex surface integrals into more manageable volume integrals.

Volume Integral

Volume integrals extend the concept of integration to three dimensions, allowing us to calculate quantities over a volume. When applying the Divergence Theorem, we move from a surface integral to a volume integral. This is because the theorem relates the flow of a vector field through a closed surface to the behavior of the field throughout the volume contained inside.
In our problem, after finding the divergence of the vector field \(\mathbf{F}\), the challenge becomes evaluating the integral over the space bounded by the paraboloid and the plane. The divergence theorem then says: \[ \iiint_{V} abla \cdot \mathbf{F} \, dV \]becomes an integral over the specific volume \(V\) described by our geometric constraints. The challenges are setting up the limits and integrating across these 3D boundaries.

Cylindrical Coordinates

Cylindrical coordinates are a coordinate system that extends the two-dimensional polar coordinates system to three dimensions by using an additional linear coordinate (often denoted \(z\) for height, alongside \(r\) for radius and \(\theta\) for angle). This system is particularly useful for surfaces or volumes that are symmetric around a central axis.
In our exercise:

• The paraboloid \(z = x^2 + y^2\) simplifies to \(z = r^2\) in cylindrical coordinates, where \(r\) is the radial distance from the z-axis.
• The plane \(z = 4\) remains as is, imposing a height boundary.

The integration limits become more intuitive, making cylindrical coordinates an ideal choice: the bounds for \(r, \theta,\) and \(z\) follow logically once the geometric shapes are expressed in terms of \(r, \theta,\) and \(z\). This simplification aids in evaluating the integral over the 3D space defined by these coordinates.