Problem 12
Question
Use the divergence theorem to find the outward flux \(\iint_{S}(\mathbf{F} \cdot \mathbf{n}) d S\) of the given vector field \(\mathbf{F}\). $$ \begin{aligned} &\mathbf{F}=15 x^{2} y \mathbf{i}+x^{2} z \mathbf{j}+y^{4} \mathbf{k} ; D \text { the region bounded by } x+y=2\\\ &z=x+y, z=3, x=0, y=0 \end{aligned} $$
Step-by-Step Solution
Verified Answer
The outward flux through surface \( S \) is the result of the volume integral of \( 30xy \) over region \( D \).
1Step 1: Understand the Divergence Theorem
The Divergence Theorem states that \( \iint_{S}(\mathbf{F} \cdot \mathbf{n}) \ d S = \iiint_{D} (abla \cdot \mathbf{F}) \ d V \), where \( \mathbf{n} \) is the outward unit normal to the surface \( S \) and \( D \) is the volume enclosed by \( S \). This transforms a surface integral into a volume integral.
2Step 2: Compute the Divergence of \( \mathbf{F} \)
Given \( \mathbf{F}=15x^{2}y\mathbf{i}+x^{2}z \mathbf{j}+y^{4} \mathbf{k} \), compute \( abla \cdot \mathbf{F} \):\( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(15x^2y) + \frac{\partial}{\partial y}(x^2z) + \frac{\partial}{\partial z}(y^4) = 30xy + 0 + 0 = 30xy \).
3Step 3: Define the Integration Limits for Region D
The region \( D \) is bounded by the planes \( x+y=2 \), \( z=x+y \), \( z=3 \), \( x=0 \), and \( y=0 \). The volume is the region between the planes \( z = x + y \) and \( z = 3 \), with base \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 2 - x \).
4Step 4: Set Up the Volume Integral
We need the volume integral \( \iiint_{D} (abla \cdot \mathbf{F}) \ dV = \iiint_{D} 30xy \ dz \ dy \ dx \). The integration limits for \( z \) will be from \( z = x + y \) to \( z = 3 \), for \( y \) from \( y = 0 \) to \( y = 2 - x \), and for \( x \) from \( x = 0 \) to \( x = 2 \).
5Step 5: Evaluate the Volume Integral
Perform the integration step by step:1. Integrate with respect to \( z \) : \[ \int_{x+y}^{3} 30xy \ dz = 30xy(z \big|_{x+y}^{3}) = 30xy(3 - (x+y)) \].2. Integrate with respect to \( y \): \[ \int_{0}^{2-x} 30xy(3-(x+y)) \ dy \]. - Expand: \( \int_{0}^{2-x} 90xy - 30x^2y - 30xy^2 \ dy \)3. Evaluate the integral with respect to \( y \): \( \left[ 45xy^2 - 15x^2y^2 - 10xy^3 \right]_{0}^{2-x} \).4. Substitute the bounds and evaluate the expression.5. Finally, integrate with respect to \( x \) from \( 0 \) to \( 2 \).
6Step 6: Compute Final Result
Substitute the upper and lower limits, simplify the polynomial, and integrate step by step, which will lead to a numerical result representing the total flux.
Key Concepts
Outward FluxVector FieldSurface IntegralVolume Integral
Outward Flux
The notion of outward flux appears when we need to calculate how much of a vector field passes through a closed surface, pointing outside of the volume it encloses.
Think of it as measuring the amount of air blowing through a balloon's surface from the inside out. The key is that every portion of the surface must be considered, and we need each tiny portion's contribution to point outward.
The outward flux of a vector field \( \mathbf{F} \) through a closed surface \( S \) can be mathematically expressed by the surface integral \( \iint_{S} (\mathbf{F} \cdot \mathbf{n}) \ dS \), where \( \mathbf{n} \) is the unit normal vector pointing outward.
To simplify the process of calculating outward flux, we can use the Divergence Theorem, which transforms this surface integral into a volume integral. This makes calculations easier when the geometry and vector field are particularly complex.
Think of it as measuring the amount of air blowing through a balloon's surface from the inside out. The key is that every portion of the surface must be considered, and we need each tiny portion's contribution to point outward.
The outward flux of a vector field \( \mathbf{F} \) through a closed surface \( S \) can be mathematically expressed by the surface integral \( \iint_{S} (\mathbf{F} \cdot \mathbf{n}) \ dS \), where \( \mathbf{n} \) is the unit normal vector pointing outward.
To simplify the process of calculating outward flux, we can use the Divergence Theorem, which transforms this surface integral into a volume integral. This makes calculations easier when the geometry and vector field are particularly complex.
Vector Field
A vector field assigns a vector to every point in a space. You can visualize it as a collection of arrows spread through an area, with direction and magnitude specific to each point.
In the given exercise, the vector field \( \mathbf{F} = 15x^2y \mathbf{i} + x^2z \mathbf{j} + y^4 \mathbf{k} \) assigns vectors based on the coordinates \( x, y, \) and \( z \). Here, each component of \( \mathbf{F} \) represents a different aspect of the field distributed in space.
The components are not random but depend on the position. For example, \( 15x^2y \) contributes to the direction along \( \mathbf{i} \) (the x-direction), and the rest depend similarly on position.
In physical scenarios, vector fields might represent various concepts such as winds on a weather map or magnetic fields around a magnet. In this problem, each position in space is influenced differently by the function of the vector field, which is critical for calculating the flux.
In the given exercise, the vector field \( \mathbf{F} = 15x^2y \mathbf{i} + x^2z \mathbf{j} + y^4 \mathbf{k} \) assigns vectors based on the coordinates \( x, y, \) and \( z \). Here, each component of \( \mathbf{F} \) represents a different aspect of the field distributed in space.
The components are not random but depend on the position. For example, \( 15x^2y \) contributes to the direction along \( \mathbf{i} \) (the x-direction), and the rest depend similarly on position.
In physical scenarios, vector fields might represent various concepts such as winds on a weather map or magnetic fields around a magnet. In this problem, each position in space is influenced differently by the function of the vector field, which is critical for calculating the flux.
Surface Integral
A surface integral helps compute the total effect of a vector field across a surface. This measurement is essential in three-dimensional spaces and particularly for closed surfaces.
The integral is expressed as \( \iint_S (\mathbf{F} \cdot \mathbf{n}) \, dS \), where \( \mathbf{F} \cdot \mathbf{n} \) denotes the dot product between the vector field \( \mathbf{F} \) and the normal vector \( \mathbf{n} \).
The term \( dS \) refers to an infinitesimal area piece on the surface.
The integral is expressed as \( \iint_S (\mathbf{F} \cdot \mathbf{n}) \, dS \), where \( \mathbf{F} \cdot \mathbf{n} \) denotes the dot product between the vector field \( \mathbf{F} \) and the normal vector \( \mathbf{n} \).
The term \( dS \) refers to an infinitesimal area piece on the surface.
- This dot product indicates how much of the field is pointing in the normal direction at each little piece of the surface.
- Adding up these contributions across the entire surface gives you the surface integral.
Volume Integral
Unlike a surface integral, a volume integral evaluates how a scalar field accumulates within a three-dimensional region.
In the context of the Divergence Theorem, it allows us to transition from considering just the surface to examining the entire volume it bounds.
The volume integral is written as \( \iiint_{D} (abla \cdot \mathbf{F}) \, dV \), where \( abla \cdot \mathbf{F} \) is the divergence of the vector field. This divergence measures the field's tendency to spread at various points.
The volume integral requires setting up limits based on the physical boundaries, calculated in the exercise using three-dimension bounds in space.
In the context of the Divergence Theorem, it allows us to transition from considering just the surface to examining the entire volume it bounds.
The volume integral is written as \( \iiint_{D} (abla \cdot \mathbf{F}) \, dV \), where \( abla \cdot \mathbf{F} \) is the divergence of the vector field. This divergence measures the field's tendency to spread at various points.
The volume integral requires setting up limits based on the physical boundaries, calculated in the exercise using three-dimension bounds in space.
- Evaluating this integral generally demands breaking it down into more manageable pieces, such as separate integrations over each dimension.
- The limits provide the range for each coordinate within the enclosed volume and are influenced by the geometric constraints given in the problem.
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