Chapter 16: Problem 42
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 \(z=3.\)
Short Answer
Expert verified
The outward flux is \(16\pi\).
Step by step solution
01
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 \(x^2 + y^2 = 25 - 3^2\).
02
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.
03
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 \(\pi \times 4^2 = 16\pi\).
04
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 \(16\pi\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
\[ \text{Flux} = \int_{S} \mathbf{F} \cdot \mathbf{n} \, dS \]
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.
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:
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.