Problem 42 - Thomas Calculus

Question

Find the outward flux of the field \(\mathbf{F}=x z \mathbf{i}+y z \mathbf{j}+\mathbf{k}\) across the surface of the upper cap cut from the solid sphere \(x^{2}+y^{2}+z^{2} \leq 25\) by the plane \(z=3.\)

Step-by-Step Solution

Answer

The outward flux is \(16\pi\).

Step 1: Examine the Surface

The sphere has a radius of 5, given by the equation \(x^2 + y^2 + z^2 \leq 25\). The surface we are examining is a cap cut by the plane \(z=3\), which is a circle centered at the z-axis with \(z = 3\) and radius \(\sqrt{25 - 9} = 4\), as \(x^2 + y^2 = 25 - 3^2\).

Step 2: Set up the Flux Integral

The flux is given by the integral of \(\mathbf{F} \cdot \mathbf{n} \, dS\), where \(\mathbf{n}\) is the outward unit normal vector to the cap. On the upper cap, the unit normal vector points upward (i.e., in the positive z-direction), so \(\mathbf{n} = \mathbf{k}\). Thus the dot product \(\mathbf{F} \cdot \mathbf{n} = 1\). The flux integral becomes \(\int_S 1 \, dS\), which simplifies to the surface area of the cap.

Step 3: Calculate the Surface Area of the Cap

The surface area of a circle can be calculated using the formula \(\pi r^2\). As found in Step 1, the radius \(r\) of the cap is 4. Therefore, the surface area is \(\pi \times 4^2 = 16\pi\).

Step 4: Conclusion

Since the vector field's normal component across the surface is 1 everywhere, the flux across the cap is simply the surface area of the cap, which is \(16\pi\).

Key Concepts

Vector Field

A vector field is a mathematical construct that assigns a vector to every point in a space. Visually, you can think of it as a map of arrows indicating direction and magnitude scattered across a region. The arrows might represent various physical quantities, like gravitational force, electric fields, or wind velocities.
In this exercise, we are dealing with the vector field \( \mathbf{F} = xz \mathbf{i} + yz \mathbf{j} + \mathbf{k} \). Here, each part of the vector field expression describes how the vector behaves with respect to the coordinate axes. Specifically:

• \( xz \mathbf{i} \) indicates the x-component, which varies with both x and z coordinates.
• \( yz \mathbf{j} \) defines the y-component that changes with y and z coordinates.
• \( \mathbf{k} \) shows a constant component in the z-direction.

These components interconnect to impact the field's behavior across the space, making it crucial to understand how each component contributes to the field's overall structure.

Surface Integral

Surface integrals allow us to calculate the accumulation of a field over a surface. They are particularly useful in physics and engineering for studying how fields like magnetism and fluid flow interact with boundaries.
When computing the surface integral of a vector field, like in our exercise, we are interested in the flux across a surface. This involves integrating the dot product of the vector field \( \mathbf{F} \) with the normal vector \( \mathbf{n} \) to the surface over the entire surface \( S \). In formula form:
\[ \text{Flux} = \int_{S} \mathbf{F} \cdot \mathbf{n} \, dS \]

• \( \mathbf{F} \cdot \mathbf{n} \) is the dot product that gives us the component of the field that is perpendicular to the surface.
• The integral sums these perpendicular components over the entire surface, effectively merging the field's interaction with the surface.

In this problem, the unit normal vector \( \mathbf{n} \) is simply \( \mathbf{k} \), as we are examining the upper cap of a sphere where the field is directed upward. The surface integral becomes a matter of calculating the surface area, which simplifies under the given conditions.

Sphere

A sphere is a perfect 3D shape where every point on its surface is equidistant from its center. The mathematical representation of a sphere is essential for understanding and solving problems related to curvilinear coordinates and surfaces.
In our exercise, the solid sphere is defined by the equation \( x^{2}+y^{2}+z^{2} \leq 25 \), which tells us that the sphere has a radius of 5 units. This implies:

• Its center is at the origin (0,0,0).
• Its radius denotes the boundary of the sphere at \( z=3 \), where its cross-section forms a circular cap.

The problem involves a spherical cap, a portion of the sphere cut by a plane. The plane cut at \( z=3 \) results in a circle with a radius of 4 units, calculated by \( \sqrt{25 - 9} \). Understanding these geometric relationships is key to visualizing and resolving the integration over this surface.