Problem 421 - Calculus Volume 3

Question

Use the divergence theorem to evaluate \(\rrbracket_{S} \mathbf{F} \cdot d \mathbf{S}\), where \(\mathbf{F}(x, y, z)=x y \mathbf{i}-\frac{1}{2} y^{2} \mathbf{j}+z \mathbf{k}\) and \(S\) is the surface consisting of three pieces: \(z=4-3 x^{2}-3 y^{2}, 1 \leq z \leq 4\) on the top; \(x^{2}+y^{2}=1,0 \leq z \leq 1\) on the sides; and \(z=0\) on the bottom.

Step-by-Step Solution

Answer

The value of the surface integral is \( \frac{5\pi}{2} \).

Step 1: Understand the Divergence Theorem

The divergence theorem states that for a vector field \( \mathbf{F} \) and a closed surface \( S \), the surface integral of \( \mathbf{F} \) over \( S \) is equal to the volume integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( S \). In formula, it is given by: \[ \int_{S} \mathbf{F} \cdot d\mathbf{S} = \int_{V} abla \cdot \mathbf{F} \, dV. \]

Step 2: Compute the Divergence of \( \mathbf{F} \)

The vector field provided is \( \mathbf{F}(x, y, z) = xy \mathbf{i} - \frac{1}{2}y^2 \mathbf{j} + z \mathbf{k} \). The divergence is computed as \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}\left(-\frac{1}{2}y^2\right) + \frac{\partial}{\partial z}(z) \). Calculating each term, we get: \[ abla \cdot \mathbf{F} = y + (-y) + 1 = 1. \]

Step 3: Define the Volume

The surface \( S \) encloses a volume \( V \) with the following boundaries: the top is the paraboloid given by \( z = 4 - 3x^2 - 3y^2 \) from \( z = 1\) to \( z = 4 \), and the bottom is the plane \( z = 0 \). The sides are the cylindrical surface \( x^2 + y^2 = 1 \).

Step 4: Set Up the Volume Integral

As the divergence \( abla \cdot \mathbf{F} = 1 \), the volume integral becomes \( \int_{V} 1 \, dV \). To evaluate this, we need the limits of integration in cylindrical coordinates where \( x = r\cos(\theta), y = r\sin(\theta) \). In cylindrical coordinates, \( z \) goes from 0 to \( 4 - 3r^2 \), \( r \) goes from 0 to 1, and \( \theta \) from 0 to \( 2\pi \).

Step 5: Evaluate the Volume Integral

The volume integral now is: \[ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{4-3r^2} 1 \, (r \, dz \, dr \, d\theta). \] First, integrate with respect to \( z \): \[ \int_{0}^{4-3r^2} 1 \, dz = 4 - 3r^2. \] Substitute this back into the double integral: \[ \int_{0}^{2\pi} \int_{0}^{1} (4 - 3r^2)r \, dr \, d\theta. \] Integrate with respect to \( r \): \[ \int_{0}^{1} (4r - 3r^3) \, dr = \left[ 2r^2 - \frac{3}{4}r^4 \right]_{0}^{1} = 2 - \frac{3}{4} = \frac{5}{4}. \] Integrate with respect to \( \theta \): \[ \int_{0}^{2\pi} \frac{5}{4} \, d\theta = \frac{5}{4} \cdot 2\pi = \frac{5\pi}{2}. \]

Step 6: Conclusion

Therefore, the surface integral \( \int_{S} \mathbf{F} \cdot d\mathbf{S} \) over the surface \( S \) is equal to \( \frac{5\pi}{2} \) according to the divergence theorem.

Key Concepts

Surface Integral

A surface integral is an extension of a line integral into any surface in three-dimensional space. Instead of integrating along a curve, a surface integral spans a surface and includes the multiplication of a vector field by an infinitesimally small vector area (a vector whose magnitude is the area of an infinitesimal piece of the surface, and whose direction is perpendicular to the surface).

For a vector field \( \mathbf{F} \), and a surface \( S \), the surface integral is denoted as \( \int_{S} \mathbf{F} \cdot d\mathbf{S} \). It calculates the flux of \( \mathbf{F} \) across the surface \( S \).

• When the flux is positive, \( \mathbf{F} \) is net outward, meaning more field lines are exiting the surface than entering.
• When the flux is negative, more of the field lines are entering than exiting, indicating a net inward direction.
• A zero value means the field lines enter and leave the surface in equal amounts.

Surface integrals are crucial in physics, especially for computing physical properties like electromagnetic flux and fluid flow across membranes in vector field contexts.

Vector Field

In mathematics and physics, a vector field assigns a vector to every point in space. A common example is the wind velocity field in meteorology, where each vector indicates wind speed and direction at each point on the map.

The given vector field, \( \mathbf{F}(x, y, z) = xy \mathbf{i} - \frac{1}{2} y^{2} \mathbf{j} + z \mathbf{k} \), describes a flow or distribution of quantities like velocity or force in three-dimensional space. The divergence of a vector field is a scalar value that provides the rate of flow expansion or contraction at each point in the field.

• The divergence is obtained by applying the divergence operator (\( abla \cdot \mathbf{F} \)) to the vector field.
• In the context of fluid dynamics, a positive divergence indicates a source or expansion, while a negative divergence denotes a sink or contraction.

Understanding the behavior of vector fields and their divergence is essential in the application of the divergence theorem.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends two-dimensional polar coordinates by adding a third axis, providing a natural means to analyze problems with cylindrical symmetry. They are especially useful when dealing with objects that exhibit rotational symmetry around a central axis.

• A point in space is described using three parameters: \( (r, \theta, z) \).
• \( r \) represents the radial distance from the origin to the projection of the point onto the \( xy \)-plane.
• \( \theta \) is the polar angle measured from the positive \( x \)-axis.
• \( z \) reflects the height above the \( xy \)-plane.

In this exercise, the surface \( S \) and the volume \( V \) are suitable for cylindrical coordinates consideration because one of the boundaries is a cylinder. With the radial distance limited to 1 by \( x^2 + y^2 = 1 \), and \( z \) ranging from 0 to a specific top boundary, cylindrical coordinates simplify the integration setup. This ensures that limits align with the object's symmetry, making calculations more straightforward.

Volume Integral

A volume integral extends the concept of an integral to three dimensions, allowing for the integration of a scalar function or a parametrically defined volume in space. It computes quantities such as mass, charge, or total divergence over a given volume.

• The volume integral of a function \( f(x, y, z) \) over volume \( V \) is given by \( \int_{V} f(x, y, z) \, dV \), which encompasses the whole volume.
• In our context, since the divergence \( abla \cdot \mathbf{F} = 1 \), the volume integral simplifies to finding the total volume \( V \) itself.
• This involves integrating the differential volume element \( r \, dz \, dr \, d\theta \) (in cylindrical coordinates), ensuring to group the appropriate limits for each variable.

The crucial advantage of using volume integrals in problems like this exercise is that they bypass the complexity of surface calculations, using the power of the divergence theorem to simplify the process. This results in efficient evaluations of complex objects in three-dimensional space.