Problem 379 - 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] \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} ; S\) is the surface of paraboloid \(z=x^{2}+y^{2}\) for \(0 \leq z \leq 9\)

Step-by-Step Solution

Answer

The value of the surface integral is \( 121.5\pi \).

Step 1: Establish Divergence Theorem

The divergence theorem states that \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds = \int_{V} (abla \cdot \mathbf{F}) \, dV \), where \( abla \cdot \mathbf{F} \) is the divergence of \( \mathbf{F} \) and \( V \) is the volume enclosed by the surface \( S \). We will use this theorem to convert the surface integral into a volume integral.

Step 2: Compute the Divergence of F

Find the divergence \( abla \cdot \mathbf{F} \). For \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), the divergence is given by \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = 3 \).

Step 3: Set Up the Volume Integral

The paraboloid described is \( z = x^2 + y^2 \) with limits \( 0 \leq z \leq 9 \). Convert to cylindrical coordinates, where \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = r^2 \). The volume \( V \) is defined by \( 0 \leq r \leq 3 \) and \( 0 \leq \theta \leq 2\pi \), with \( 0 \leq z \leq 9 \).

Step 4: Integrate Over the Volume

Evaluate the volume integral using cylindrical coordinates: \[ \int_{0}^{2\pi} \int_{0}^{3} \int_{0}^{r^2} 3r \, dz \, dr \, d\theta \].First, integrate with respect to \( z \):\[ \int_{0}^{r^2} 3r \, dz = 3r[z]_{0}^{r^2} = 3r^3 \].Next, integrate with respect to \( r \):\[ \int_{0}^{3} 3r^3 \, dr = \frac{3}{4}[r^4]_{0}^{3} = \frac{3}{4}(81) = 60.75 \].Finally, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} 60.75 \, d\theta = 60.75 \times 2\pi = 121.5\pi \].

Step 5: Conclusion

Thus, the value of the surface integral \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds \) is \( 121.5\pi \).

Key Concepts

Surface Integral

Surface integrals are an important concept in vector calculus and are used to calculate the flow of a vector field across a surface. When dealing with a vector field \( \mathbf{F} \) and a surface \( S \), the surface integral of \( \mathbf{F} \) over \( S \) represents the total effect of the vector field over that surface. This can be thought of as computing how much of the field is `passing through' the surface.

• Surface Integrals are denoted by \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds \), where \( \mathbf{n} \) is the outward unit normal vector.
• This integral sums up the dot products of the field \( \mathbf{F} \) with the normal vectors over the surface.
• It essentially calculates a weighted sum of \( \mathbf{F} \)'s components normal to \( S \).

Understanding the surface integral is crucial when applying the Divergence Theorem, which allows conversion between surface integrals and volume integrals, simplifying complex calculations.

Vector Fields

A vector field assigns a vector to every point in a space. Think of it like giving each location an arrow depicting a force, velocity, or any vector quantity.In our example, \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) is a simple, linear vector field where:

• The x-component is represented by \( x \mathbf{i} \).
• The y-component is represented by \( y \mathbf{j} \).
• The z-component is represented by \( z \mathbf{k} \).

The divergence of a vector field, denoted \( abla \cdot \mathbf{F} \), measures the tendency of the field to originate from or converge into a point. It is calculated as the sum of the partial derivatives of the components of \( \mathbf{F} \).

• For this particular vector field, the divergence is a constant value \( 3 \), indicating a uniform "source strength" throughout the space.

This makes our calculations using the Divergence Theorem straightforward, as the integral of a constant over a volume is simply the product of the constant and the volume.

Cylindrical Coordinates

Cylindrical coordinates are a 3D coordinate system which extends polar coordinates by adding a height (z) dimension. They are particularly useful for problems involving symmetry around an axis, like our paraboloid, making calculations simpler compared to Cartesian coordinates.In cylindrical coordinates, a point in space is described by:

• \( r \), the radial distance from the z-axis.
• \( \theta \), the angle in the plane from the positive x-axis.
• \( z \), the height above or below the xy-plane.

To shift from Cartesian to cylindrical coordinates, we use these transformations:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( z = z \)

In the given exercise, using cylindrical coordinates allows the paraboloid \( z = x^2 + y^2 \) to be expressed simply as \( z = r^2 \). This formulates a set boundary in terms of \( r \) and \( \theta \), enabling straightforward integration over the volume for the 3D region enclosed by the surface.