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Magazine Chapter 19: Problem 54
Calculate the flux of the vector field through the surface. \(\vec{F}=e^{y^{2}+z^{2}} \vec{i}\) through the disk of radius 2 in the \(y z-\) plane, centered at the origin and oriented in the positive \(x\) -direction.
Short Answer
Expert verified
The flux through the disk is \( \pi (e^4 - 1) \).
Step by step solution
01
Understand the Problem
We need to calculate the flux of the vector field \( \vec{F} = e^{y^2 + z^2} \vec{i} \) through a disk of radius 2 in the \( yz \)-plane. The disk is centered at the origin \((0,0,0)\) and is oriented in the positive \( x \)-direction.
02
Define the Surface
The surface of the disk lies in the \( yz \)-plane, and points in the positive \( x \)-direction. The parametric representation of the disk can be expressed as \((x, y, z) = (0, y, z)\) where \(y^2 + z^2 \leq 4\).
03
Determine the Normal Vector
Since the surface is in the \( yz \)-plane and oriented in the positive \( x \)-direction, the normal vector is \( \vec{n} = \vec{i} = (1, 0, 0) \).
04
Compute the Flux Integral
The flux of a vector field \( \vec{F} \) through a surface \( S \) with a normal \( \vec{n} \) is given by \( \iint_S \vec{F} \cdot d\vec{S} = \iint_S \vec{F} \cdot \vec{n} \, dA \). Here, \( d\vec{S} = \vec{n} \, dA \), with \( dA \) being an infinitesimal area element on the surface.
05
Evaluate the Dot Product
Determine \( \vec{F} \cdot \vec{n} \). Since \( \vec{F} = e^{y^2 + z^2} \vec{i} \) and \( \vec{n} = \vec{i} \), we have \( \vec{F} \cdot \vec{n} = e^{y^2 + z^2} \cdot 1 = e^{y^2 + z^2} \).
06
Set Up the Integral
The integral becomes \( \iint_{S} e^{y^2 + z^2} \, dA \), over the region where \( y^2 + z^2 \leq 4 \). This region describes a circle of radius 2 on the \( yz \)-plane.
07
Convert to Polar Coordinates
Convert the region to polar coordinates: \( y = r \cos \theta \), \( z = r \sin \theta \), with \( dA = r \, dr \, d\theta \). The limits for \( r \) are from 0 to 2, and for \( \theta \) are from 0 to \( 2\pi \).
08
Perform the Integration
The integral in polar coordinates is \( \int_{0}^{2\pi} \int_{0}^{2} e^{r^2} r \, dr \, d\theta \). Substitute \( u = r^2 \), \( du = 2r \, dr \), giving \( \frac{1}{2} \int_{0}^{2\pi} \int_{0}^{4} e^{u} \, du \, d\theta \).
09
Integrate and Simplify
Compute the integrals: \( \frac{1}{2} \times 2\pi \times \left[ e^{u} \right]_0^4 = \pi \times \left( e^4 - 1 \right) \). Thus, the flux is \( \pi (e^4 - 1) \).
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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 where each point in space is associated with a vector. In simple terms, imagine a space populated with arrows, each pointing in a specified direction and having specified magnitude. These arrows represent the vector at each point in that field. The vector field
- Can be visualized in three-dimensions where each location in space ( \(x, y, z\)) is tied to a vector compensated either by direction or intensity.
- May represent various physical phenomena such as electromagnetic fields, gravitational fields, or fluid flow in physics and engineering.
- Is given by functions like \(\vec{F} = e^{y^2 + z^2} \vec{i}\), where here every point of the field is associated with the direction of the x-axis and magnitude dependent on \(y\) and \(z\).
Surface Integral
A surface integral is a method used to compute a type of integration over a surface in three-dimensional space. Think of it as measuring how a field interacts with or penetrates a surface. In a practical sense:
- The surface integral extends the concept of line integrals to surfaces, allowing the evaluation of flux or circulation of a field over a surface.
- Typically, involves combining the field function, like our vector field, with an infinitesimal surface element \(d\vec{S}\), often expressed with a normal vector.
- When calculating the flux through a surface using a surface integral, consider the formula: \(\iint_S \vec{F} \cdot d\vec{S} = \iint_S \vec{F} \cdot \vec{n} \, dA\), where \(\vec{n}\) is the normal vector to the surface.
Polar Coordinates
Polar coordinates offer an alternative to Cartesian coordinates when dealing with problems involving circles or disks, like the surface integral through a disk in our problem. These are particularly handy because:
- They simplify calculations involving circular or cylindrical shapes by converting from the standard \((x, y)\) coordinates to \((r, \theta)\), where \(r\) is the radius and \(\theta\) is the angle.
- In conversion, we use \(y = r \cos \theta\) and \(z = r \sin \theta\), which effectively allows us to express areas in terms of \(r\) and \(\theta\).
- The area element \(dA\) on the surface becomes \(r \, dr \, d\theta\), making integration over circular surfaces simpler and more direct.
Normal Vector
A normal vector is a vector that is perpendicular to a surface at a given point. Its importance in computing flux lies in its role defining the orientation of a surface.
- For surfaces in three-dimensional space, the normal vector helps specify the direction in which the surface is facing, which is crucial for calculations involving vector fields.
- In our example, since the disk is in the \(yz\)-plane oriented in the positive \(x\)-direction, the normal vector is \(\vec{n} = \vec{i} = (1, 0, 0)\).
- The proper determination of the normal vector impacts the accuracy of the surface integral, as it must correctly represent the interaction between the vector field and surface.
